VP v VJ ,0 Cforenion (prcee |S>«ue ELEMENTARY PLANE TRIGONOMETRY NIXON Ronton HENRY FROWDE Oxford University Press Warehouse Amen Corner, E.C. Qtew $orft 112 Fourth Avenue £farenfcon (|>reee Retries ELEMENTARY PLANE TRIGONOMETRY that is Plane Trigonometry without Imaginaries by & R. C. J. NIXON, M.A. AUTHOR OF 'EUCLID REVISED,' 'GEOMETRY IN SPACE, ETC. BOOTOH COLLEGE LIBRARY " CHESTNUT HILL, MASS. MATH. DEPT. AT THE CLARENDON PRESS 1892 PRINTED AT THE CLARENDON PRESS BY HORACE HART, PRINTER TO THE UNIVERSITY 150577 PREFACE It is customary to define an elementary course of trigonometry as Trigonometry to the end of solution of Triangles. But this defi- nition is obviously very vague, and settles neither what of the subject is included nor what is excluded. Casting about for some natural line of demarcation between the regions of elementary and higher trigonometry, it occurred to me that such a line is given by the use or non-use of the symbol V — i . Thus Elementary Trigonometry may be appropriately defined as that part of the subject which can be done without the use of imaginaries ; while the remainder, which requires imaginaries, will then constitute Higher Trigonometry. The present book treats of Elementary Trigonometry, defined as above ; and the points proposed in its composition are — i°, to include only what specially concerns the Trigonometrical Functions — excluding all that seems more properly to appertain to Arithmetic, Algebra, Mensuration, or Pure Geometry; hence, for example, the theory of logarithms is omitted, as strictly belonging to treatises on Algebra : the only exceptions to this rule are that a few extraneous matters, which seemed indispensable to the subject, yet could not be well disposed of by reference to text-books, are treated of in a Preliminary Chapter ; and that, for the same reasons, a Note on Convergence precedes the treatment of series. 2°, to utilize the gain of space thus acquired for a more thorough and complete amplification than is usual of what does specially concern the Trigonometrical Functions. vi Preface 3°, to give all definitions and proofs in their fullest generality, and with the strictest ac curacy from the first ; my experience being entirely against provisional and limited definitions or proofs, fol- lowed afterwards by more complete extensions. 4°, to work out in full a large number of specimen Examples — 150 of these are given — and to add hints to many of the Exercises. In the arrangement of the order of treatment, and in the details of the proofs, I have not followed, or even been guided by, any previous treatise ; but have relied on my own judgment, based on the experience gained in twenty-five years' labour as a teacher of the subject. Exact logical accuracy of sequence and method has been everywhere aimed at; and the rule followed has been to introduce everything when, and not until, it is needed, and then to treat it completely. Of course a subject of this kind cannot be learned without studying it in books ; but, in writi?ig this work, I have had very little direct help from any books, unless it be from the two Trigonometries of the late Professor De Morgan. For the unique accuracy and thoroughness of all De Morgan's writings I feel unbounded admiration, and to them I am largely indebted. There have lately appeared some four or five Trigonometries which, judging from their advertisements, are of an elaborate character. To avoid the possibility of being influenced by them in any way, I have avoided even looking at these ; nor have I any idea what they contain or do not contain. Indeed any other course with regard to contemporaneous works seems to be quite unjustifiable. For an author to make himself acquainted with the modern books with which he is, by the nature of the case, in direct competition, may be a very ingenious way of surpassing them ; but the proceeding is, to say the least, questionable. The whole book was read, both in MS. and 'proof,' by pupils of mine at the Belfast Institution ; and I relied on them, almost ex- clusively, to aid me in correcting press errors. They also worked Preface vii a large number — from a half to three-quarters — of the 1200 Exer- cises. I have thus, as the book was in progress, been in constant touch with the exact class for whom it is intended — the upper boys at schools, who are preparing to compete for University Scholar- ships. My grateful acknowledgements are due to — • i°, the Rev. J. Milne (of Dulwich) for leave to take what I liked from his Companion : this kind liberality was very useful to me in the chapter wherein the Lemoine and Brocard circles are briefly treated of. 2 , Professor Purser (of Queen's College, Belfast) for his in- genious elementary proof of Gregories Series, printed on pages 316 to 318: this proof has been just recently invented by him, and is now here first published. 3 , Mr. R. Chartres (of Manchester) for his original Theorem, printed on page 330, of which he has given me the copyright: by means of this very useful Theorem, the summation of numbers of series, hitherto considered to require the use of V—i, can be effected by elementary trigonometry. Some other new proofs sent to me are acknowledged in their respective places. The late Professor Wolstenholme not only gave me permission to make any use I pleased of his Problems, but also, when I sent him a list of those which I proposed to extract, most kindly re- turned me answers and corrections in all cases where such were needed : every Problem taken is indicated in its place. Of the other Exercises many are taken from the Educational Times' Reprints — references to the volumes being added — a few are private, and the rest come from various public sources, fre- quently indicated. Those to which the letters Q. C. B. are appended were set by Professor Purser. In the subsidiary matters of type, diagrams and pagination, Vlll Preface neither expense nor trouble has been spared to make the work perspicuous. In regard to the settlement of the details connected with these, I have to thank the authorities of the Clarendon Press for full permission to have my own ideas carried out in their integrity. As far as practicable the pages have been arranged so that a diagram and its corresponding text may be on view together, with- out turning a page : see, for example, pages 48 to 63. Attention to such details adds greatly to a Student's ease and comfort in studying. I hope some day to supplement this Elementary Trigonometry by a small companion volume to be called De Moivre's Theorem and its Consequences. R. C. J. NIXON Royal Academical Institution, Belfast May, 1892 CONTENTS including formulae which should be known PRELIMINARY— Angles— Functions— Graphs Def's of the units (degree, radian) of angle Circumf O = 2 7T r . Area „ — it r 2 . Arc „ = r0 . Sector „ = J r 2 Two right angles = 77 radians . x/180 = 6/ir A radian = 5 7°- 2 95 78 nearly One right angle = 90 = 1-5708 of a radian To eight places of decimals 7T = 3.14159265 22 355 Approximations for it are 7 ' 113' 3.1416 CHAPTER I — The Trigonometrical Functions Def's of the T.F.s Each T.F. in terms of each other . sin 2 CX + cos 2 a = 1 . tan 2 OK + 1 = sec 2 oc 1 + cot 2 a = cosec 2 a tan OC = sin OC /cos CX If tan CX m m , then sinCX = n Vm 2 + n 2 , and cos CX = PAGK 1-2 3 4 5 and 280 6 9 9 9 11, 12 12 M-34 24 28 29 V 30 nrr + n- X Contents CHAPTER II— Possible values of the Trigonometrical Functions ......... 35-47 Values of T. F.s for o°, 90 , 180 , 270 , 360 .... 37 Successive changes of value of T. F.s 38-41 Graphs of T. F.s . . 4 2 ~47 CHAPTER III — On certain relations between the Trigono- metrical Functions of Angles whose sum or difference is a multiple of a Right Angle 48-57 sin OC = cos (90 — a) = sin (i8o°-a) = - sin (-a) = -cos (9o° + a;> = -sin (i8o° + a) cosa = sin(9o°-a) = -cos (i8o°-a) = cos(-a)= sin(9o° + a) = -cos(i8o°+a) CHAPTER IV — Fundamental relations between Trigono- metrical Functions of Angles and Trigonometrical Functions of parts of those Angles . sin (a + ft) = sin OC cos ft + cos OC sin ft cos (OC + ft) = cosa cos/3 - sin OC sin ft sin (OC - (3) = sin acos/3 - cos OC sin ft cos (OC - (3) = cos OC cos (3 + sin a sin ft sin 2 OC = 2 sin OC cos OC . cos 2 OC = cos 2 a - sin 2 a = 2 cos 2 a — 1 = 1- 2 sin 2 a sin 2 n a = 2 n sin a cos a cos 2 a cos 4 a .. sin (OC + ft) + sin (OC -ft) = 2 sin a cos /3 sin (a + /3) - sin (OC - (3) = 2COsOCs\nft cos (OC + ft) + cos (OC- ft) = 2 cos a cos ft cos (a + ft)- cos (a - /3) = - 2 sin OC sin /3 cos 2 n_1 a sina + sin/3= 2 sin — cos a + /3 . sin a — sin /3 = 2 cos — sin OC- ft 2 a- ft 5«-94 58 }■> 60 >) 64 >; 65 Contents xi a+/3 a- & . COS(X+ COS/3 = 2 COS COS . 65 2 2 a + /3 . oc - /3 cosft — cos/3 = — 2 sin sin . „ ^ 22 tan a ± tan 8 tan (« + (3) = -^ „. a .... ^ i + tan (Xtan/3 _ + cot a cot /3 - I cot (OC + 8) = — — T ^ro~ ... 66 v - ^ cot oc ± cot /3 2 tan OC tan 2 a = cot 2 oc i — tan 2 a cot 2 oc — i tan {OC + 8 + y) = kttt wi ; ... 67 ^ ^ /J 1 — 2 (tan (3 2 cot a Stan a - ITtana tan y) sin 3 a = 3 sin a - 4 sin 3 a ... „ cos 3 OC = 4 cos 3 a — 3 cos CX ... „ . 1 — cos 2 a so tan 2 a= .... 68 1 + cos 2 OC 1 — tan 2 a cos 2 a = t — 5 — „ 1 + tan 2 a 2 tan OC sin 2 oc = 1 + tan 2 a sin (OC + (3) sin (a - /3) = sin 2 a - sin 2 /3 . . . . „ cos (oc + /3) cos (oc — /3) = cos 2 a + cos 2 /3 — 1 . . . 69 a + /3 . ol - 8 sin a sin j3 = sin 2 sin 2 . ,, ' 2 2 oc + 8 o oc - 8 cos OC cos ^ = cos 2 + cos 2 I . j, ^2 2 at - c 3 t 3 + c 5 ts- ... tan n OC = — =Hb =-iz » where t = tan OC 70 1 - C 2 t 2 + C 4 t 4 - ... v n r\ n - 2 , r* V n - 4 cotn(X = , „ x = cot OC „ C v n-l r* v n-3 , c v-n-5 1 X — O3 X + V^5 X — . . . xii Contents PAG P. If s = sin OC and c = cos OC sin 5 CX = 16 s 5 — 20 s 3 + 5 s .... 72 cos 5 a = 16 c 5 - 20 c 3 + 5 c . ... „ 2 sin OC = + Vi +. sin 2OC ± Vi — sin 2 a . . 73 2 cos a = + Vi + sin 2 a + Vi — sin 2 a . . „ If A + B + C = 180 2 sin A = 4]! cos — 7= 2 /\ 2 cos A r= 4lTsin — +1 2 " 2sin2A = 4 IT sin A 2 cos 2 A = - 4 IT cos A - 1 76 2 tan A = ITtanA ** 2 (tan - tan — ) = 2 cot A = II cot A + ITcosecA . 78 2 cos 2 A = 1 - 2 FT cos A 7g If OC + (3 + y = 360 2 cos 2 oc = 1 + 2 IT cos oc >> CHAPTER V— Numerical values of the Trigonometrical Functions of some special angles 95-115 sin 45 = — = cos 45 9 - v 2 sin 30 = i = cos6o° 97 COS30 = — SI = sin 6o° . 2 " sin 15° = I (V6 - V2) IOO cos 15°= 1 ( 170 xiv Contents When 6 < t:J 2, tan 6 - 6 > 6 - sin When 6 so small that 4 is negligible i To sin = 6 3 6 2 COS0 = i 2 tan0 = 6 3 + — 3 174-175 cos 1/ = .9999999 .... 178 sin 10"= .000048481 .... 179 When n }> 10, sin n" = nsin i' „ CHAPTER X — The Theory of Proportional Differences . 185-200 Theory not true for sines, tangents, secants when OC is near 77/2 .......... 186-189 nor true for cosines, cotangents, cosecants when OC is near zero ,, nor true for tan OC when OC is between 77/4 and 7T/2 . . 188 „ „ cot OC „ „ „ o „ 77/4 . , „ nor true for logarithmic T, F.s when OC is near zero or 7r/2 191-192 CHAPTER XI— Relations between Sides and Angles of a Triangle , 201-220 a/sin A = b/sin B = c/sin C = D . . . 201-202 a = b cos C + c cos B . . . . 203 sin A = — -v^ssj s 2 s 3 ..... 204 cosA= (b 2 + c 2 -a 2 )/ (2 be) . . . 205 A = \ be sin A = Vss, s 2 s 3 = \ V^S b 2 c 2 - 2a* = a 2 sin B sin C/ (2 sin A) ,, Contents xv PAGE sin - = V %^ 206 2 be tan 2 A /ss t cos — = v — J . 2 be B-C / cot - = (b - c)/(b + c) . . .207 and 354 .. . A / B-C A / . (b + c) sin — / cos = a = (b — c) cos — / sin B-C 2 CHAPTER XII— On the adaptation of formulae to Loga- rithmic Calculation , . 221-226 CHAPTER XIII— Solution of Triangles .... 227-240 Given 3 sides , 228-229 Given 2 sides and the included angle . . . . 230 Given 2 angles and a side ....... 231 Given 2 sides and an angle opposite one of them — the ' Ambiguous Case ' . 231-233 Small errors ...,.,,., 238-239 CHAPTER XIV — Lines and Circles remarkably associated with the Triangle 241-272 A / Internal bisector of A = 2 be cos — / (b + c) , . 242 External do = 2 be sin — /fb'vC in — /(b* >> Median = J Vb 2 + c 2 + 2 be cos A . 243 Symmedian = -r-x 5 x median ... ,, b 2 + c 2 Sides pedal A are a cos A, b cos B, c cos C . . . 244 Angles „ 180 - 2 A, 180 - 2 B, 180 - 2 C . 245 Area „ 2 A IT cos A XVI Contents R = abc/(4 A) = a/(2 sin A) . , A . B . C / A r = A/s = Si tan — = asm — sin — / cos — ' 1 2 2 2 / 2 a/ A B C / A r\ = A/St = s tan — = a cos — cos — / cos — ' J 2 2 2 / 2 r = 4 RIT sin — T 2 „ . A B C r, = 4 R sin — cos — cos — 2 2 2 r x + r 2 + r 3 - r = 4 R . . A = Vrr x r 2 r 3 . s 2 = 2 (r x r 2 ) . A = R x semi-perim' pedal A In-radius pedal A = 2 R IT cos A cot co = 2 cot A = 2 a 2 /(4 A) . cosec 2 co = 2 cosec 2 A R] (rad' of T. R. G) = i R sec co . R 2 (rad' of cosine G) = R tan co R 3 (rad' of Taylor 0) = R VII sin 2 A + II cos 2 A CHAPTEE XV— Quadrilaterals and other Polygons . Def of a negative area ...... Area of any convex quad' ...... PAGE 251 252 253 254 255 256 266 1} 268 273-286 273 274 = ho - d) (0- - e) (cr - f) (cr - g) - defg cos 2 Area of any cross quad' 7*}' 275 - V(tr- d - e) (0- - d - f) (0- - d - g) + defg sin 2 0-4>)i r Area at any quad' = ^ xy sin co where cos co = {(d 2 <%. e 2 ) + (f 2 ^ g 2 )}/(2 xy) 276 Contents xvii PAGE Circum-radius convex quad' = £ a / (de + fg) (df + eg) (dg + ef) V (o- - d) (o- - e) (o- - f) ( 1 is pos' T 33 22 4X^3X3 2Xj 2 X 1 X 2 X 3 Sx 2 136 12 tan (77/7 + <£), &c...tan (w/7 + <£>), -tan (ttA-c/)), &c „ ...6 from end + — T 49 17 2(R-r) 2 (R + r) 169 14 64800 648000 „ ...last but one it OL 183 7 cotC cosC 193... Example 1 19" 41" „ ... Exercise 2 9-59°95 T 9-59Q935 1 „ ... Exercise 4 9-657056 9-9657056 211. ..last but one Hyparchus Hipparchus 220 last c' C 221 last sec (OL-][ 4i7504 = 3-J4I632I = 3-1416025 C 20 48 = 3-i4i595i C 40 96 = 3- I 4i5933 V^Of.fl '256 '512 '1024 These agree to 5 decimal places ; and, as the O is intermediate in area to the two pol's, we see that the formula- area of a O = 3-14159 x (its radius) 2 is a close approximation. Hence 7T = 3-14159 very nearly. i2 Preliminary We shall habitually use 3-1416 as a good and sufficient ap- proximation *. Note (1) — It is often useful to remember that 3 141 6 = 3 x 7 x8 xn X17. Note (2) — If the calculations be continued further, the areas of the polygons will be found to agree up to the figures 3-14159265 ; so that in a circle whose radius is a mile, the error in area, by taking these figures for 77, is less than the ten-millionth of a square mile ; that is less than a square foot. The value of 77 to 8 decimal places may be recollected by the following jingle— "A 77 receipt may thus be given; "From 3 millions take 8 thousand seven; "Increase remainder 5 per cent; "Result to 8 decimals represent." 3000000 8007 2991993 J 49599 6 5 3.14159265 — 77 to 8 places. Note (3) — The following practical f rules are worth noting. Since 22/7=3.142857 = A 2 say .'. A x x -0004 = -001257142857 = C x „ .-. Aj-Ci = 3.1416 exactly = A 2 „ Again C x x -006 = .00000754 = C 2 „ •'• A 2 - C 2 = 3.14159245 = A 3 „ A 3 is correct to 6 decimal places, i.e. as correct as 355/113. Hence we get the following rules to multiply by 77 : As a 1st approx' use 22/7 for 77. As a 1st correction subtract -0004 of the product from itself. * The fraction 22/7 (originally found by Archimedes, and correct to two decimal places) is often used as a rough approximation for 77. The fraction 355/113 (originally found by Metius, and correct to six decimal places) is a very close approximation to 77. To recollect this last fraction put down each of the first three odd numbers twice, giving 1 13355 ; then the three right hand figures are the numerator, and the three left hand are the denominator. For an account of these approximations see an article by Dr. Glaisher in the Messenger of Mathematics \ New Series, II. 123. + Taken from Milne's ' Companion] by permission of the Editor. Angles — Functions — Graphs 13 This gives a 2nd approx' as true as by using 3-1416 for 77. As a 2nd correction subtract -006 of the 1st correction from the 2nd approx'. This gives a 3rd approx' true to 6 places; and is .\ as true as by using 35 5/ 1 1 3 for 77. Similarly may be verified the following rules for approximately dividing by 77: i°, multiply by 7/22 ; 2 , add -0004 of this result to itself; 3 , add -006 of this last result to itself. The final result is true for six places. All questions of conversion, or reduction, of angles from one measure to another, evidently belong to the domain of simple Arithmetic. So also questions concerning areas and arcs of circles, or sectors of circles, belong to the province of Mensuration. Hence all such questions are foreign to Trigonometry properly so-called : however, to render this Preliminary Chapter more complete, we add here some Examples and Exercises of this character. Examples 1. To express a radian in terms of degrees, minutes, and seconds. If x° = one radian x 1 then -5- = — 180 77 •'• x = ^TT^ = 57-29578, nearly 3.1416 i. e. a radian = 57°- 2 9578 = 57 17' 44"-8, nearly. Note— It will be found that this = 57°^ - (■£/- (£)" to within (-ji_) 2. To express a degree in terms of a radian. If T 6 = one degree then - = -— r 77 180 n 3-1416 , .-. 6 = Z = ' OI 745> nearl y 1 00 i. e. a degree = -01745 of a radian, nearly. 1 _\// 14 Preliminary 3. To express 23 14' 57" in centesimal units. 23 14' 57"= 23 i4 / -95 I (dividing by 60 mentally in = 2 3 . 2 49 1 6 j each case) = (^ x 23-24916)^ = 252.8324073 = 25s 8 3 v 2 4 ".o73 4. To express 23214*57" in degrees and sexagesimal units. 23 g i4 x 57"= 23S.1457 = (A x 23-1457)° = 20°.83i3 = 20 49 / -8678 \ (multiplying by 60 ment- or? 20 49' 52"-o68 j ally in^each case) 5. If a centesimal minute is taken as the unit of angular measure, what represents a sexagesimal minute ? i'= Gfr)°« C¥ x eV) s = (inr) x = a 7 a )* i. e. a sexagesimal minute is represented by -f f . 6. /^7zatf length of rope will enable an animal, tethered by it, to graze over an acre of ground? In a question like this, where rough approximation is evidently sufficient, we may take ~- for tt. Let x yds. be the length of rope. Then 22 x 2 = 4840 .-. x 2 = 1540 whence x = 39-24 i. e. 39^ yds. is very nearly the length req'd. 7. What is the length of a column which, at the distance of a mile, subtends an angle of\°at the eye ? Owing to the smallness of the A > and the great distance of the column as compared to its length, we may consider the column as an arc of a of radius 1760 yds. r - r 3 I 4 1 ^ -I .-. length of column = 1760 x -7: yds. s ' 180 x io 4 3 = 3o-7!78 i. e. the column is nearly 31 yds. high. Angles — Functions— Graphs 15 8. Show that if the circumference of a circle whose radius is ioo miles is calculated by using 2>5B/ II 3f or 7T> th e error is less than 4 inches. Show also that the areal error, on the same supposition, is less than 2 acres. By division 355/113 = 3-14159292 &c But 77 = 3-14159265 &c .". the error, by taking 355/113 for 77, ^> -0000003 .*. error in circumf (which = 200 77) ^> -00006 of a mile i.e. j> 3-8016 inches and .*. < 4 inches. Again areal error ^> -0000003 x 10000 sq' miles 3> -003 x 640 acres ^1-92 acres and .*. < 2 acres. Exercises 1. Express 77/13 radians in degrees, and decimals of a degree. 2. Express 13 as a decimal of a radian. 3. Find approximately the radian measure of the angle which contains as many degrees as its supplement contains grades. J oh' Camb': / 8o. 4. If the unit angle subtends an arc equal to the diameter, how will one- third of a right angle be expressed ? 5. A train is going OL miles an hour on an arc of a circle of (3 miles radius : how many seconds of angle will it turn through in n seconds of time ? 6. What is the length of an object, which at the distance of a mile, subtends an angle of one minute ? 7. Find (in degrees, minutes, and seconds) the angle which at the centre of a circle of 8 feet diameter, subtends an arc of 10 feet length. 8. At what distance would a man 6 feet high subtend an angle of one minute ? 9. If the radius of a circle is 20 inches, find the radii of three concentric circles, by which the original circle is quadrisected. 10. If the measures of the angles of a triangle, referred to i°, ioo', ioooo // , respectively, as units, are in the proportion of 2 to 1 to 3, find the angles. Math' Tri'\ '73. 1 1 . Find the length of an arc on the sea which subtends an angle of one minute at the centre of the Earth, supposing the Earth a sphere of diameter 7920 miles. Z. U. ''84. 16 Preliminary 12. If the distance between the centres of the Earth and the Moon is sixty times the radius of the Earth, find the angle which the radius of the Earth sub- tends at the Moon's centre. Trin' Camb': '44. 13. The apparent diameter of the Moon is 30' : find how far from the eye a circular plate of 6 inches diameter must be placed so as just to hide the Moon. Trin' Camb': '49. 14. If two plumb lines, suspended from points at a fixed distance apart, are inclined to each other at a small angle of m" when at the Earth's surface ; and at an angle of x\" when at an elevation h ; show that the Earth's radius is nh/(m — n) nearly. Pet' Camf: '50. 15. If the number of units of measure in any angle is equal to the number of degrees in that angle less the number of radians in it ; how many degrees are there in the unit of measurement ? 16. Calculate it to two decimal places on the assumption that an angle of 200 subtends at the centre of a circle an arc which is 3-^- times the radius. Oxf Jun' Locals: / 86. 17. Calculate in inches, correct to four decimal places, the length of an arc of a circle which with a radius of a mile subtends an angle of one second at the centre. Ox? Sen' Locals : '86. 18. Show that with every unit the numbers expressing an angle of an equi- lateral triangle, and an angle of a regular hexagon, would together double the number expressing an angle of a square. Joh' Camb' : '76. 19. Assuming that the Earth is a sphere whose diameter is 7912 miles, show that (to three decimal places) the length of an arc on its surface, subtending a degree at its centre, is 69-045 miles ; and that this result will be obtained whether we use 3-1416 or 3-1415926 for 71". 20. Calculate 77 (to four decimal places) from the following data — In a circle, whose radius is 10 feet, an arc of 3 feet nf inches subtends an angle of 22 30' at the centre. 21. If the diameter of a bicycle is 50 inches, show that the number of revo- lutions it makes in 9 seconds is very nearly the same as the number of miles per hour at which it is going. (Take 22/7 for 77) 22. Assuming that the Earth moves in a circle whose radius is 95 000 000 miles, and that a year is 365 days exactly ; show that its rate of motion is nearly 19 miles per second. 23. Three equal circles (each of one foot radius) are so placed that each touches the other two ; find (to three decimal places of square feet) the curvi- linear area enclosed between the three circles. Angles — Functions — Graphs 17 24. Find (to three decimal places of square inches) the area of a segment of a circle (whose radius is one inch) cut off by a chord equal to the radius. Q. C. B. '74 25- Calculate the velocity of light from the following data — The Earth s diameter is 7900 miles ; the angle this diameter subtends at the Sun is i7"'8 ; the time the Sun's light takes to reach the Earth is 8 m I3 S> 3. Q. C. B. '71. 26. Calculate the rate of the Moon's motion, in miles per hour, from the following data — The distance between the centres of the Earth and Moon is 59-964 times the Earth's radius ; the time of the Moon's revolution round the Earth is 27 days, 7 hours, 43 minutes, II seconds ; the Earth's radius is 3963 miles. Q. C. B. '81. 27. Calculate, so that the error shall be less than the /1000th part of the answer, the weight in grains of a globe of standard gold, whose radius is one metre ; given that — Volume of a * sphere is f77 (radius) 3 ; specific gravity of standard gold is 17-157 ; a cubic inch of water weighs 252-46 grains ; a metre is 39-37 inches. Q. C. B. '83. 28. If the Earth's orbit is assumed to be a circle whose diameter is 184 000 000 miles, find the area enclosed by this orbit. What error is caused by taking 3-1416 instead of 3-1415926 for 77? 29. It is known of a pair of regular polygons that the ratio of the number of grades in an angle of one to the number of radians in an angle of the other is 160 to 77 : show that the pair must be one of four. Joh' Camb' : '78. 30. When the unit of angular measure is the angle whose arc is 77 times the radius, the angles of a triangle are such that the number of degrees in one, the number of grades in another, and the number of units of angular measure in the third are all equal : find each angle in radian measure. Pet' CamV\ 66. 31. Of the angles of a triangle one contains as many grades as another contains degrees, and the third contains as many centesimal seconds as there are sexagesimal seconds in the sum of the other two : find the radian measure of each angle. Joh' Camb' \ '72. 32. The apparent angular diameter of the Sun is half a degree : a planet is seen to cross its disc in a straight line at a distance from the centre equal to three-fifths of the radius : prove that the angle subtended at the Earth by the part of the planet's path projected on the Sun is 77/450. Math' Tri'\ '74. * See Geometry in Space : p. 69. C 18 Preliminary 33. ABC is a triangle such that, if each of its angles in succession is taken as the unit of measurement, and the measures formed of the sums of the other two, these measures are in arithmetical progression : show that the angles of the triangle are in harmonical progression. Math' Tri'\ '79. 34. Show that there are eleven, and only eleven, pairs of regular polygons such that the number of degrees in an angle of the one is equal to the number of grades in an angle of the other ; and that there are only four pairs when these angles are expressed in integers. Find also the number of sides of each. Math' Tri'\ '66 and '81. Note — Ifx, y are the numbers of sides, cond'ns of question give the indeter- minate efn xy + 18 x — 20 y = o, or (x — 20) (y + 18) + 360 = o. 35. For what pairs of regular polygons is it true that the number of degrees in each angle of the one is to the number of radians in each angle of the other as 144 to IT ? 36. Given, in two regular polygons, the ratio of the number of sides in one to the number of sides in the other ; and given also the ratio of the number of degrees in an angle of one to the number of grades in an angle of the other ; find the number of sides in each. 37. One angle of a triangle is x degrees, another is x grades, and the third is x radians ; find, approximately, the magnitude of each angle in degrees, minutes, and seconds. 38. Two chords of a circle (radius 3 feet) intercept arcs 77/3 and 77/5 : find (in degrees) the angle between the chords. Ox' Jun' Math' Schol': '74. 39. Two circles cross each other orthogonally: if d is the distance between their centres ; and if their radii are as ^3 to 1 ; find (in terms of d) the area common to the circles. As we have already, to some extent, seen in Algebra and Geometry, it is necessary and in the highest degree convenient, for purposes of mathematical generalization, to consider that if any process is conventionally taken as positive, then (what may be loosely called) the reverse process is to be taken as negative. Examples The positive processes — receipt of money ; motion to the right ; motion upwards ; rotation like the hands of a clock — have as their respective negative analogues — payment of money ; motion to the left ; motion down- wards ; rotation opposite to that of the hands of a clock. Angles — Functions — Graphs J 9 Def — If two straight lines of unlimited length are taken at right angles to each other, they divide the plane (indefinitely extended) which contains them, into four parts ; and each of these parts is called a quadrant. Note — Of course, strictly speaking, a quadrant is the fourth part of a circle : the foregoing definition maybe brought within this usual meaning by supposing a circle whose centre is the cross of the lines, and whose radius is of indefinite length. In Trigonometry — and thence in mathematics generally- following conventions are made. -the X' o V- X Def— Let XOX',YOY' be two straight lines at right angles to each other, and indefinitely ex- tended — where XOX'is drawn in direction from the right hand towards the left; and YOY' is drawn in direction from the top of the paper towards the bottom: then — Y f i°, the area separated off by OX, OY is called the first quadrant; » „ „ OY, OX' „ second „ OX',OY' „ third „ OY', OX „ fourth „ 2°, lines measured in direction X'OX are called positive ; and therefore „ „ „ XOX' „ negative. „ „ „ Y'OY ,, positive; and therefore „ „ „ YOY' „ negative. 3°, angles measured by rotation about O in an anti-clock-wise direction (that is following the cycle XYX^) are called positive ; and therefore those measured by a clock-wise rotation (that is fol- lowing the cycle XY r X'Y) are called negative. C 2 20 Preliminary Examples If OPj, 0P 2 , 0P 3 , 0P 4 are respectively in the 1st, 2nd, 3rd, and 4th quadrants ; and PiMi, P 2 M 2 ,&c. J_sonXOX , ; P1N1 , P3N3 , &c. _L S on YOY'; then OM l5 OM 4 , ON l5 ON 2 are considered ^s-z'/zV ; and .-. OM 2 , OM 3 , ON 3 , 0N 4 „ „ negative. Also, if OP^ OP 2 , &c. are imagined to have revolved, from coincidence with OX, in an anti-clock-wise direction, the A s XOP 1? XOP 2 , XOP 3 , XOP 4 are considered positive ; and .\ if they are imagined to have revolved in a clock-wise direction, the A s XOP 4 , XOP 3 , XOP 2 , XOP x are considered negative. Def'—M XOX r , YOY' are two fixed lines at right angles, and PM, PN the respective distances of any point P, in their plane, from them ; then PM, PN are called the rect- angular coordinates of P with respect to the rectangular axes XOX', YOY', and the origin O : also O M (to which PN is equal) is called the abscissa of P, and PM is called the ordinate of P. Note— It is usual to denote OM, PM respectively by such pairs of letters as (x, y) or (h, k) or (a, b). Evidently the position of P, with respect to the axes, is completely determined when its coordinates are known. N P M X' IS N H X H V Angles — Functions — Graphs 21 Def — If O is a fixed point and OX a fixed direction, and if P is any point, then OP and the angle POX are called the polar coordinates of P with respect to the pole O, and initial line OX; also OP is called the radius vector of P, and POX the vectorial angle of P. Note — It is usual to denote OP, and POX, respectively, by such pairs of letters as (r, 6) or (p, /a 2 — x 2 * See Magntis'' Hydrostatics : 3rd ed', p. 120. Angles — Functions — Graphs 23 is represented by a circle. But again other graphs are found to go off to an infinitely removed distance : such e. g. is the case with the graph of y = ^/^Z^ For the satisfactory tracing of some of these, certain subsidiary lines called asymptotes are requisite. Def — If as a curve is indefinitely extended it continuously approaches to (without attaining within any finite distance) coin- cidence with a fixed straight line, so that it may be considered to touch that line at an infinitely removed distance along the line, then the straight line is called an asymptote to the curve. For further information on the subject of Graphs the Student is referred to Chapter XV of Professor Chrystal's Algebra, where he will find diagrams and discussions of several graphs. A large number of excellent Examples of the application of graph-drawing to practical questions in Political Economy will be found in Professor Marshall's Principles of Economics. Algebraic Note Sarrus' practical rule to write down the expansion of the three column oc j3 y determinant OC' 13" y" Repeat the 1st two col's, and connect by diag / s thus- X Put down the product of each set of 3 constituents lying along a diag 7 line; prefixing + if the line goes from the top towards the right, and — if from the top towards the left: this gives as result a/3'y" + /3/ a" + ya'/3" - jSa'y" - ay'/3" - y/3'a" The above very useful rule is not usually given in Algebras ; and .'. has been printed here for reference. CHAPTER I The Trigonometrical Functions N X X f J G Y N X P \ V \A § |. Def's— Let XOX', YOY' be X lines, in which OX, OY are pos' direct. Let OA, initially along OX, revolve about O either way round, in the plane of XOY, thro' an angle 06. Then OA must be in one of the 4 quadrants into which the pi' is divided by XOX', YOY'; and .*. must assume a position such as one of the 4 indicated in the fig's. In OA take any pi P; and drop PN ± on XOX'. The sides of the A PON will afford three ratios and their reciprocals : of these — NP OP ON the ratio ^5 is called 'the sine of 06' — usually written sin 06 cosine ,, „ „ cos 06 tangent „ „ „ tan 06 „ cotangent „ ,, ,, cot 06 OP NP ON ON NP 55 >> )5 )) 5 j j: 55 55 The Trigonometrical Functions 25 . OP the ratio ^— — is called : the secant of OL ' — usually written sec 06 ON OP NP cosecant „ „ ,, cosec 06 The expression 1 — cos 06 is sometimes called the versed- sine of 06, and is then written vers 06; but this function is not much used. Def — The triangle PON, constructed as above for the angle 06, is called the auxiliary triangle of 06 ; and the ratios of the sides of the auxiliary triangle, when indicated by the names given above, are called *the trigonometrical functions of the angle 06. Note (1) — Of the lines which are the terms of the foregoing ratios, the revolving line OP is considered to remain invariably positive in all positions ; while the other lines ON, NP follow the usual conventions, as to sign, explained in the Preliminary Chapter. Note (2) — We shall use the contraction T. F. for the words 'trigonometrical function.' Note (3) — From the mode of its construction, the auxiliary A is (for any definite A) always of the same species, no matter where in OA the p't P is taken ; and .\ the T. F.s are invariable for any (the same) A« Note (4) — The ratios defined as T. F.s depend only on the relative positions of the initial and revolving lines, and .-. will be the same for all A s in which these lines occupy the same relative positions. Now the addition (positive or negative) of any multiple of 4 right A s to an A will make no change in the relative positions of the lines forming that A ' hence — If any A = 0t° = *$, and/ (CX) or f {&) denotes any T. F. of that A, then /(a) =/(m. 360 + a) and f(6) =/(m. 2TT + 6) where m is any integer pos', neg', or zero. Hence we see that the T. F.s are periodic ; and that the extent of one of their periods is 2 it. It will be seen hereafter that the tangent and cotangent have a shorter period, viz. IT. * Sometimes also the circular functions, and sometimes the goniomefrical ratios of the angle. l6 Trigonometry — Chapter I § 2. The sine has been denned as NP/OP {not PN/OP) and hence from def we see that — for A s between o° and 180 the sine is pos'; while 3J 3? nesr 180 „ 360 Sim'ly for the other functions ; so that the following table is at once apparent from the def 's. A between sin cos tan cot sec cosec + o° and 90 + + + + + 90 and 180 + — — — — + 180 and 270 — — + + — — 270° and 360 — + — — + — A Note — The auxiliary A PON, with respect to an OC, having been constructed (as above) then, calling PN its perp? (perpendicular) ON its base, and OP its hyp' (hypotenuse) the T. F.s are sometimes denned thus — sin OC = perpyhyp', cos OC = base /hyp', tan OC = perp'/base We do not recommend this mode of definition v — i°, it takes no account of the sign of the lines NP, ON ; and, 2°, it is apt to give an inadequate (or even inaccurate) conception of the T. F.s for A s greater than a/t A- § 3. By def 7 , cot 06, sec Oi, cosec Oi are respectively reciprocals of tan Oi, cos Oi, sin a, so that sin Oi. cosec Oi = 1 cos Oi. sec a = 1 tan Oi . cot 06 = 1 The Trigonometrical Functions 27 Def — Here we first come on what are called trigonometrical identities : that is relations between the functions which are true for all values of the angle (or angles) involved There is a large number of trigonometrical identities which must be recollected, if any success is to be attained in the applications of trigonometry. Those which are imperatively necessary to be committed to memory will be printed for reference in a list at the end of the book. § 4". To express any one T. F. [call it the first) in terms of any other one (call it the second) follow this plan — i°, construct fig's as in § 1 : 2 , write down the def of the 1st: 3 , if of the three sides of the auxil' A one occurs in the 1st but not in the 2nd, eliminate that side by means of the relationship between the sides given by Enc' i. 47 : 4 , divide each term of the fraction thus obtained by the den'r of the 2nd: 5 , replace the ratios by their T. F. equivalents. The process will then be complete. Example To express cos OC in terms of cosec OC. i°, take the fig's of § i 2°, COSCX ON ~ OP r, >/OP 2 -NP 2 OP ^(S)"- 4 > OP NP Vcosec 2 OC — 1 h j cosec OC M M M 8 1 8 8 1 1 8 <3 o 23 o 7^ o CO O o o CO O o o CO 1—1 1 8 8 d (M o (-1 o o CO CO co "> SI o CO 8 M o CO o CO o co o o h- 1 M + + 8 8 « l« hi +-> M "M +-• o ■+-> o o o o I o o i o "> V o o M + 8 +-■ o o o o 1 *> + o o 8 c cc 8 c CO + I M [8 1 + 8 d (M 'N a CO CO 1— 1 £2 -M +-> CO + + +-> M M "> "> CO o o "> CO o o 8 8 "> "> 8 c "co 8 c. "co 8 "co "> 8 8 1 c3 » where A is some const', and A n = cos OC) Square and add ; then A 2 (m 2 + n 2 ) = sin 2 OC + cos 2 OC = i .-. A = sin OC — Vm 2 + n 2 m Vm 2 + n 2 n - as before. and cos OC = - Vm 2 + r The Trigonometrical Functions 31 Exercises 2 X f X ■+■ I ") 1. If tan OC = -, find sin OC and cos OC. 2 x + 1 2. Given that tan OC = i, find each of the other T. F.s. 3. If cot OC = — , find sin OC and cos OC. q 4. If sin OC = 1, find the value of cos OC + cot OC + cosec OC. 5. Given that sin OC — -012, find each of the other T. F.s to three decimal places. \^2 — 1 6. If vers OC = =r- , find the value of */2 sin OC + cos OC + tan OC + cot OC + sec OC + cosec OC. f\ / 2 2 \ 3 7. If tan 3 (f) = — , show that OC cosec (f) + j3 sec (f) = \0C^ + fi*r- 8. Express each of the T. F.s of OC in terms of vers OC. 9. If cos

ON, unless they coincide, and OP > NP, „ „ i, • ; but ON, NP may have any relative values; we see that sin 06 > 1, and cos 06 > 1 ; also that sec 06 <£ 1, and cosec 06 < 1 ; but that tan a and cot 06 are unlimited in magnitude. 3» Trigonometry— Chapter II Examples 1. The equation sin OC = x + -is impossible, ifx is real. For if y = x + - X then x 2 — xy + i = o .-. 2X = y + Vy 2 — 4 .*. , for real values of x, the least value of y is 2 i i. e. x + - o, unless x = y x 2 + y 2 > 2xy, „ ■■■ (x + y) 2 > 4xy „ 4xy (x + y) 2 < i, ■ 4xy '"' secarF (xT7y 2 " " § 10. We are now in a position to trace the gradual changes in the values of each T. F., as the angle changes from zero to four right angles. To trace the changes of mag7iitude and sign in sin OL, as OL changes continuously from o° to 360 . X' N Let XOX', YOY' be rect'r axes; where OX, OY are pos' direct. Let O P, initially along OX, revolve anti-clock-wise A thro' an OL. Drop PN Ion XOX'. Possible values of the T. F.s 39 NP Then sin 06 = -^-p i°, when 06 = o, sin 06 = o. As OP rotates thro' the 1st quadrant XOY, NP increases con- tinuously; until, when 06 = 90°, NP coalesces with OP along OY: also NP remains pos 7 thro 7 the quadrant. .'. , as 06 changes from o° to 90 , sin 06 changes continuously from o to 1, rem'g pos 7 all the way. 2 , when 06 = 90 , sin 06 = 1. As OP rotates thro' the 2nd quadrant Y OX', NP diminishes continuously; until, when 06 = 180 , NP vanishes: also NP remains pos 7 thro 7 the quadrant. /. , as 06 changes from 90 to 180°, sin 06 changes continuously from 1 to o, rem 7 g pos 7 all the way. 3 , when 06 = i8o° ; sin 06 = o. As OP rotates thn/ the 3rd quadrant X 7 OY 7 , NP increases continuously; until, when 06 = 270 , NP coalesces with OP along OY 7 : also NP, having changed sign from pos 7 to neg 7 as it passed thro' the value o, remains neg' thro 7 the quadrant. .*. , as 06 changes from 180 to 270°, sin 06 changes continuously from o to — 1 , ren/g neg 7 all the way. 4 , when 06 = 270 , sin 06 = — 1. As OP rotates thro 7 the 4th quadrant Y'OX, NP diminishes continuously; until, when 06 = 360°, NP vanishes: also NP remains neg 7 thro 7 the quadrant. .*. , as 06 changes from 27o°to 360°, sin 06 changes continuously from — i to o, rem 7 g neg 7 all the way. To trace the changes of magnitude and sign in COS 06, as 06 changes continuously from o° to 360°. Construct as for the sine. ON Then cos 06 = ^-~ i°, when 06 = o, cos 06 = 1. 40 Trigonometry— Chapter II As OP rotates thro' the ist quadrant XOY, ON diminishes continuously; until, when 06 = 90°, ON vanishes: also ON remains pos' thro' the quadrant. .*. , as 06 changes from o° to 90°, cos 06 changes continuously from 1 to o, reir/g pos' all the way. 2 , when 06 = 90 , COS 06 = o. As OP rotates thro' the 2nd quadrant YOX', ON increases continuously; until, when 06= 180°, ON coalesces with OP along OX': also ON, having changed sign from pos' to neg' in passing thro' the value o, remains neg' thro' the quadrant. .*. , as 06 changes from 90° to 180°, cos 06 changes continuously from o to —1, rem'g neg' all the way. 3 , when 06 = 180°, COS 06 = — 1. As OP rotates thro' the 3rd quadrant X'OY', ON diminishes continuously; until, when 06 = 270°, ON vanishes: also ON remains neg 7 thro' the quadrant. .-. , as 06 changes from 18b to 270°, cos 06 changes con- tinuously from — 1 to o, rem'g neg' all the way. 40, when 06 = 270°, COS 06 = o. As OP rotates thro' the 4th quadrant Y'OX, ON increases continuously; until, when 06 = 360°, ON coalesces with OP along OX: also ON, having changed sign from neg' to pos' in passing thro' the value o, remains pos' thro' the quadrant. .*. , as 06 changes from 270° to 360 , cos 06 increases con- tinuously from o to 1, rem'g pos' all the way. To trace the changes of magnitude and sign in tan 06, as 06 changes continuously from 0° to 360 . Construct as for the sine. NP Then tan 06 = ^^v ON i<>, when 06 = o, tan 06 = o. As OP rotates thro' the ist quadrant XOY, NP increases and ON diminishes (both continuously) until, as 06 approaches 90°, Possible values of the T. F.s 4 1 ON diminishes indefinitely; so that, ultimately, when 06 = 90 , the ratio N P/O N becomes indefinitely large : also both N P and O N remain pos' thro' the quadrant. .*. , as 06 changes from o° to 90 , tan 06 changes continuously from o to 00 , rem'g pos 7 all the way. 2 , when 06 = 90 , tan 06 = 00 . As OP rotates thro' the 2nd quadrant Y OX', NP diminishes and ON increases (both continuously) until, when 06 = 180 , NP vanishes; so that, ultimately, when 06 = 180 , the ratio N P/O N vanishes : also N P remains pos' thro' the quadrant ; but ON, which changed sign from pos' to neg' in passing thro' the value o, remains neg' thro' the quadrant. .'. , as 06 changes from 90 to 180 . tan 06 changes continuously from 00 to o, rem'g neg' all the way. 3 , when 06 = 180 , tan 06 = o. As OP rotates thro' the 3rd quadrant X'OY', NP increases and ON diminishes (both continuously) until, as 06 approaches 270 , ON diminishes indefinitely; so that, ultimately, when 06 = 2*70°, the ratio N P/O N becomes indefinitely large : also O N remains neg' thro' the quadrant ; and N P which changed sign from pos' to neg' in passing thro' the value o, remains neg' thro' the quadrant. .*. , as 06 changes from 180 to 270 , tan 06 changes continuously from otooo, rem'g pos' all the way. 4 , when 06 = 270 , tan 06 = 00 . As OP rotates thro' the 4th quadrant Y'OX, NP diminishes and ON increases (both continuously) until, when 06 = 360 , NP vanishes; so that, ultimately, when 06 = 360 , the ratio N P/O N vanishes : also N P remains neg' thro' the quadrant ; but ON, which changed sign from neg' to pos' in passing thro' the value o, remains pos' thro' the quadrant. .'. , as Oi changes from 2 7o°to 360 , tan 06 changes continuously from 00 to o, rem'g neg' all the way. Note — The Student should now find no difficulty in tracing the changes of the other T. F.s 42 Trigonometry— Chapter II § II. The graphs of the different T. F.s afford excellent aid towards recollecting their successive changes. These we now pro- ceed to construct roughly : to draw them with any great accuracy requires a knowledge (which we shall get in Chapter V) of the numerical values of the T. F.s of several angles intermediate between those which are exact multiples of a right angle. To draw the Graph of sin 06 — sometimes called the Sinusoid; and sometimes the Curve of Sines. Take recfr axes; and let OX, OY be their pos' direc's. Let A 15 A 2 , A 3 , A 4 be p'ts in OX, such that U A x = Aj A 2 = A 2 A 3 = A 3 A 4 ; and let each of these equal lines represent a r't A . At A x draw A x B 15 J_ to OX, and in pos' direc'; and at A 3 draw A 3 C 3 , -L to OX, and in neg' direc'; so that each of them is a unit of length. Then the graph of sin 06 will be a curve drawn from O thro' B 15 A 2 , C 3 , A 4 ; and if P is any p r t on the graph, and PN is J_ to OX, then PN represents sin 06, for that value of 06 represented by ON. Note — The filling in of the curve between the above 4 p r ts wall be more easily seen when the numerical values of a few intermediate sines have been calculated. E.g. (in § 31) sin 30 is found to be \, so that when ON = |OA 1? PN should be \ A a B r Also, for the easier practical drawing of graphs, the Student should use what is sometimes called logarithmic paper ; viz. paper Possible values of the T. F.s 43 ruled in two _L direc's, with the rulings rather close together— say /ioth inch apart. The absolute lengths of OAj and A x B x are arbitrary, and need not necessarily bear any particular relation to each other ; but there is a certain convenience in taking them so that OA x : A x B 1 = 3 : 2 ; for then, inasmuch as 3 : 2 = 90 : 6o°, and a radian is not far from 6o°, it will happen that the line which represents the sine, for a particular A, will also nearly represent that angle in radian measure. This would enable us to draw the graph of a function of 6 and sin a, where r = 0i°. Some Exercises in graph drawing will be given after Chapter V. The remarks in this Note are to be taken also in connection with each of the following graphs. To draw the Graph of COS OL — sometimes called the Cosixmsoid. Take rect'r axes; and let OX, OY be their pos' direc's. Let A 15 A 2 , A 3 , A 4 be p'ts in OX, such that vj Aj = r\ 1 A 2 — A 2 A 3 = A 3 M 4 , and let each of these equal lines represent ar't A • In OY take B, so that OB is a unit of length; at A 2 draw A 2 C 2 in the neg' direc r ; and at A 4 draw A 4 B 4 in the pos 7 direc', so that A 2 C 2 = A 4 B 4 = OB. Then the graph of cos OL will be a curve drawn from B thro 7 A lf C 2 , A 3 , B 4 ; and, if P is any p r t on the graph, and PN is _L to OX, then PN represents cos a, for that value of 06 represented by ON. 44 Trigonometry — Chapter II To draw the Graph of tan Oi. Take rect'r axes; and let OX, OY be their pos' direc's. Let A 1} A 2 , A 3 , A 4 be p'ts in OX, such that UAj = A x A 2 = A 2 A 3 = A 3 A 4 ; and let each of these equal lines represent a r't A . Thro' A x and A 3 draw indefinitely extended lines B x A x C a , B 3 A 3 C 3 , _L to OX ; where A x B l; A 3 B 3 are in pos' direc', but A x C 13 A 3 C 3 are in neg' direc'. Then the graph of tan Oi is a series of 3 curves drawn thus — i°, from O in the pos' direc' so as to have A 2 B x an asymptote ; Possible values of the T. F.s 45 2°, thro' A 2 in the neg' direc' so as to have A x C x an asymptote, and in the pos' direc' so as to have A 3 B 3 an asymptote ; 3°, from A 4 in the neg' direc' so as to have A 3 C 3 an asymptote ; and if P is any p't on the graph, and PN ± to OX, then PN represents tan 06 for that value of OL represented by ON. Sim'ly the graphs of the other 3 T. F.s can be drawn : we give the diagrams, leaving the Student to follow out the details as above. Notice that all 3 have asymptotes. 4 6 Trigonometry — Chapter II Bx B, C, A, A* X Graph of sec Oi Possible values of the T. F.s 47 Graph of cosec OL CHAPTER III On certain relations between the Trigonometrical Functions of Angles whose sum or difference is a multiple of a Right Angle § 12. Theorem — If any angle is 06°, then sin 06 = cos (90 — 06) and COS 06 = sin (90 — 06) >G Take XOX 7 , YOY' ± lines, where OX, OY are pos 7 direct. Let O A, initially along OX, revolve in pos 7 direc 7 (i. e. anti-clock- wise) thro 7 06°. To construct 90 — 06; let OB, initially along OX, revolve in pos 7 direc 7 thro' 90 (this will bring it to OY) and then back, in neg 7 direc 7 , thro 7 06°. Then, from the mode of construction, we see that When O A is in the 1st quadrant, OB is in the 1st 2nd „ , „ „ 4 th 3 rd » > „ ,, 3 rd 4th „ , „ „ 2nd In OA take any p 7 t P, and drop PM _L on XOX 7 . „ OB „ „ Q, „ „ QN „ T. F.s of Complementary Angles 49 Then, since MOP = YOB = NQO, A A and M = N ; /. A s MOP, NQO are sim'r. Also, by examining the fig's, we see that MP and ON are always of the same sign. _- . MP , ON , .*. the ratios -^-p and 07s nave always the same sign and mag- nitude. .•". sin 06 = cos (90 — a) o 1 , OM _ NQ . bo also the ratios -^-p- and -^-?s nave always the same sign and mag'. .-. cos 06 = sin (90 — 06) Hence, if any two A s are complementary — the sine of either = the cosine of the other Sim'ly, or as a deduction from the foregoing, we should get that, if any two A s are complementary — the tangent of either = the cotangent of the other and „ secant „ „ „ cosecant „ „ Note — It is from these properties that are derived the words Co-sine, equivalent to sine of Complement, Co-tangent, „ „ tangent „ „ , and Co-secant, „ „ secant ,, ,, Examples A B + C If A + B + C = i 8o°, then — and are complementary ; , , ■ A B + C so that then sin — = cos 2 2 A . B + C and cos — = sin 2 2 E 5o Trigonometry — Chapter Ml §13. Theorem — If any angle is QL°, then sin a = sin (180 — a) and COS 06 = — COS (180 — 06) M N X X r N N Y X X r \ / \B W Q M Y' Take XOX 7 , YOY 7 -L lines, where OX, OY are pos 7 direct. Let OA, initially along OX, revolve in the pos 7 direc 7 (i. e. anti- clock-wise) thro 7 06°. To construct 180 — 06; let OB, initially along OX, revolve in pos 7 direc 7 thro 7 i8o° (this will bring it to OX 7 ) and then back, in the neg 7 direc 7 , thro 7 06°. Then, from the mode of construction, we see that when OA is in the ist quadrant, OB is in the 2nd 2nd „ , „ „ ist 3 rd » > » » 4th j 3 ,-, ?3 4th „ , „ „ 3rd In OA take any p 7 t P, and drop PM ± on XOX'. „ OB „ „ Q, „ „ QN „ „ A A Then, since POM = QON, A A and M = N ; .-. A s POM, QON are sim'r. Also, by examining the fig 7 s, we see that MP and NQ are always of the same sign. T. F.s of Supplementary Angles 51 MP _ NQ t .*. the ratios np and ^— ^ have always the same sign and magnitude. .-. sin a = sin (180 — oc) Again OM and ON have always opposite signs, so that always OM ON OP ~~ OQ .-. COS 06 = — cos (180 — 06) Sim'ly, or as a deduction from the foregoing, we should get that tan a = — tan (180° — oc) COt 06 = — COt (l8o° — Oi) sec 06 = — sec (180 — 06) cosec 06 = cosec (180 — 06) So that the T, F. of any A is expressible in terms of the same T. F. of the supplementary A . Examples sin 120 = sin (180 — 120 ) = sin 6o° cos 150°= — cos (180 — 150 ) = — cos 30 vers (180 - oc) = 1 - cos (180 - a) = 1 + cos a Exercises Simplify each of the following — 1. sin 90 + tan 2 (180 - oc) - cosec 2 (90 - a) 2. cos (90 - a) sin (180 - oc) - sin (90 - OC) cos (180 - oc) 3. vers (90 - OC) vers (180 - OC) B + C A 4. cos (B + C) - cos cos (180 — A) + sin- where A, B, C are angles of a triangle. E 2 52 Trigonometry — Chapter II! § 14- . Theorem — If any angle is 06°, then sin a = — sin (i8o° + a) and COS a = — COS(i8o° + 06) N x r a b/ Q / A A V MX X'MO Y' N M Y r X X' Q\ /P B Ay /B B\ a/ \ N X X r N O M p \. Take XOX 7 , YOY 7 , J. lines, where OX, OY are pos 7 direc 7 s. Let OA, initially along OX, revolve in the pos 7 direc 7 (i.e. anti- clock-wise) thro 7 06°. To construct i8o° + 06; let OB, initially along OX, revolve in the pos 7 direc' thro' 06° (this will bring it to OA) and then on in the pos 7 direc 7 thro 7 i8o°. Then, from the mode of construction, we see that when O A is in the ist quadrant, OB is in the 3rd „ „ „ 2nd „ , „ „ 4th }) jj >j 3*^- 5? 3 11 11 ist » „ „ 4 tn „ , „ „ 2nd In OA take any p't P, and drop PM J_ on XOX'. „ OB „ „ Q, „ „ QN „ „ Since AOB is a st 7 line, A s POM, QON are clearly sim 7 r. Also MP and NQ are always of opposite sign. , ,. MP , NQ .'. the ratios -^-^ and ^r^z are always of the same magnitude, OP but of opposite sign OQ sin a = — sin (180 + a) T. F.s of (i8o° + a) 53 So also the ratios ^p and ^-p are always of the same mag', but of opposite sign. .*. cos Oi = — cos (i8o° + a) Sim'ly, or as a deduction from the foregoing, we should get that tan oi = tan (180 + a) cot a = cot (180 + a) sec Oi = — sec (180 + a) cosec a = -- cosec (180 + Oi) Note — We saw (on p. 25) that all the T. F.s are periodic, and that 360 (or 2 77) is the extent of period for each of them. But we can now find shorter periods for the tangent and cotangent. For, since tan OC = tan (180 + OC) it follows that 180 may be added any number of times to OC, without altering the value of tan Oi. Also, anticipating the result of § 16, we have tan (— 180 + Oi), which is the same as tan { — (180 — 00} = - tan (180 - Oi), by § 16 = tan a, by § 13 So that — 180 may be added any number of times to OC, without altering the value of tan OC. .-. tan OC = tan (n . 180 + OC) where n is any integer, pos', neg', or zero. Sin/ly cot OC = cot (n . 180 + OC) Expressing these important results in radian measure, we have tan 6 = tan (n it + 0) cot = cot (mr + 6) Hence the periods of the tangent and cotangent are 180 (or it). Examples tan 915 = tan (5 x 180 + 15 ) = tan 15 cot (- 885 ) = cot {5 x (- 180 ) + 1 5 } = cot 15 54 Trigonometry— Chapter III §15. Theorem — If any angle is 06°, then sin 06 = — cos (90 + a) and COS 06 = sin (90 + 06) B N Q X' N O P^A N M X X' M B / Q M N Q N Y' X X' o \ Q, / B M ^6 B Take XOX 7 , YOY 7 J_ lines, where OX, OY are pos 7 direc's. Let OA, initially along OX, revolve in the pos 7 direc 7 (i.e. anti- clock-wise) thro 7 06°. To construct 90 + 06; let OB, initially along OX, revolve in the pos 7 direc 7 thro 7 90 (this will bring it to OY) and then on, in the pos 7 direc 7 , thro 7 06°. Then, from the mode of construction, we see that when O A is in the 1st quadrant, OB is in the 2nd 2nd ., , „ „ 3rd 55 55 55 3^ ?' 5 55 55 4^" 5? 55 55 4^ J, , ,, ,, ISt In OA take any p 7 t P, and drop PM _L on XOX 7 . „ OB „ 5, Q, „ „ QN „ „ Then, since POM = YOQ = OQN, A and M A N .'. A s POM, OQN are sim'r. Also, by examining the flg 7 s, we see that M P and O N are always opposite in sign. T. F.s of (90 + oi) 55 MP , ON u , , • /. the ratios -^= and -—- have the same magnitude but oppo- site signs. .-. sin oc = — cos(9o° + a) OM NQ So also the ratios -^-^ and -^r have always the same sign and mag'. .-. cos 06 = sin (90 + a) Sim'ly, or as a deduction from the foregoing, we should get that tan 06 = - cot (90 + 06) cot 06 = — tan (90 + a) sec 06 = cosec (90 + a) cosec 06 = — sec (90 + 06) Note — We know that the removal of any multiple of 4 right A s from an A will leave any T. F. of it unaltered ; and, by the foregoing results, we see that 2 right A s can be removed, or the supplementary A substituted, if, in certain cases, a change of sign is made. Hence the T. F.s of large A s can be reduced to the same T. F.s of smaller A s - Again, one right A can be removed, or the complementary A substituted, if we change the T. F. into its complementary T. F.s, and, in certain cases, make a change of sign. Examples Since 48 f = 360 + 180 - 53 .'-. sin 487° = sin (180 - 53°) = sin 53 Again, since 1365 = 3 X360 + 180 + 90 + 15 \ cos 1365 = cos (180 + 90 + 1 5 ) = — cos (90 + 15°) = sin 15 Exercises Simplify each of the following — 1. vers (90 + oc) vers (90 — OC) + vers-CX vers (180 — a) 2. sin 903 + sin 643 56 Trigonometry— Chapter III §16. Theorem — If any angle is 0L°, then sin a = — sin (— a) and COS 0i = COS (—a) Take XOX', YOY / X lines, where OX, OY are pos' direc's. Let OA, OB, initially along OX, revolve respectively in pos' and neg' direc's, thro' 06°. Then, from the mode of construction, we see that when OA is in the ist quadrant, OB is in the 4th, 55 >> j' 2nd ,, , ,, ,, 3^5 35 51 j> 3^ >> 5 35 >> 2nu, 3? >> 3? 4^n ,, , j, ,, isl. In OA take any p't P, and drop PN _L on XOX', producing it to meet OB in Q. Then obviously — 1°, AONPE A ONQ; and, 2 , NP, NQ are always of opposite sign. NP NQ .-. the ratios -^-p and -=-= are always equal in magnitude and opposite in sign. /. sin oi = — sin (— Oi) c 1 u ON . ON _ bo also the ratios ^r-^ and ^-^ have always the same sign and mag 7 . OP — OQ COS Oi — cos ( — oc) T. F.s of (-06) 57 Sim'ly, or as a deduction from the foregoing, we should get that tan a = — tan (— a) cot a = — cot (— a) sec a = sec(— 06) cosec 06 = — cosec (—06) § 17. Summing up all the preceding results, we have sin 06= cos (90 — 06) = sin(i8o° — a) = — sin (— 06) = — cos (90 + 06) = - sin (180 + 06) cos 06 — sin (90 — a) = — cos (180° — a) = cos (— 06) = sin (90 + a) = — cos (180 + a) If a° = T 0, then these formulae, in radian measure, are sin 6 = cos (71-/2 — 0) — sin (77- — 6) = — sin (— 6) = — cos (77-/2 + 6) = — sin (ir + 6) cos 6 = sin (77-/2 — 6) = — cos (77- — 6) = cos (— 6) = sin (77-/2 + 0) = — cos (77- + 6) From these the analogous formulae for the other T. F.s are at once deducible. Note — The Learner should make himself familiar with the above results: they are of the utmost importance ; for, by means of them the T. F. of any A can be expressed in terms of the same T.F. of a pos' A less than 90 . In case of forgetting any one of them, it can be easily recovered by considering its particular fig' in the simple case when OC < 90 . All such simple cases are here included in the adjoining diagram, which should be carefully studied. CHAPTER IV Fundamental relations between Trigonometrical^ Functions of Angles and Trigonometrical Functions of parts of those Angles § 18. Theorem — If QL and /3 denote any angles, then sin (06 + /3) = sin 06 cos /3 + cos 06 sin /3 and cos (06 + /3) = cos 06 cos /3 — sin 06 sin /3 Take the case when 06, /3 are pos' A s , and each of the three 06, /3, 06 + /3, < 9 o°. Let a line, initially along OX, revolve in the pbs' direc' (i.e. anti- clock-wise) thro' 06 into position OA, and then on thro 7 (3 into position OB. Take P any p't in OB: drop J_s PM on OX, PQ on OA, QN on OX, QR on PM. Now Then MP OP QPR = 9 o° - PQR = RQO = 06. NQ + PR NQ OQ PR PQ "T Again OP "OQ OP 1 PQ OP sin (06 + /3) = sin a cos /3 + cos 06 sin /3 OM _ ON - RQ ON OQ RQ QP OP - OP "OQOP PQ'OP cos (a + /3) = cos a cos (3 — sin 06 sin /3 Expansions of sin (06 + /3) and cos (a + /3) 59 Now sin (90 + a + /3) = cos (a + /3) = cos 06 cos f3 — sin 06 sin /3 = sin (90 + 06) cos /3 + cos (90 + oc) sin /3 And cos (90 + 06 + /3) = — sin (06 + (3) = — sin 06 cos /3 — cos 06 sin /3 = cos (90 + a) cos /3 — sin (90 + 06) sin /3 .*. the restriction that 06 < 90 may be removed. Sim'ly „ „ „ /3 „ „ „ .'. the theorem is true for all pos' A s . Lastly, if 06 = — x, and /3 = — y, then sin (06 + /3) = sin {— (x + y)} = — sin (x + y) = — sin x cos y — cos x sin y = sin (— x) cos (— y) + cos (— x) sin (— y) = sin oc cos /3 + cos 06 sin /3 and cos (06 + /3) = cos { — (x + y)} = cos (x + y) = cos x cos y — sin x sin y = cos (— x) cos (— y) — sin (— x) sin (— y) = cos 06 cos /3 — sin 06 sin /3 .-. the theorem is true for all neg' A s ; and, .*. , is true for all values of 06 and f3. Note— 'By (OC + (3) is denoted a single angle, where OC, (3 are any angles such that (3 added to OC gives its magnitude. For example : sin 13a = sin (8a + 5a) = sin 80C cos 5a + cos 8asin 5a 6o Trigonometry — Chapter IV § 19. Theorem — IfOL and ft are any angles, then sin (a — ft) = sin a cos ft — cos a sin ft #« ft. Let a line, initially along OX, revolve in the pos' direc' (i. e. anti- clock-wise) thro' 06 into position OA, and then back, in the neg' X direct thro 7 ft into position OB. Take P any p't in OB: drop J_s PM n OX, PQ on OA, QN on OX, PR on QN. Now Again Then PQR = 90 - OQN = 06 MP NQ - QR NQ OQ QR QP OP " OP = OQ'OP QPOP sin (a — /3) = sin 06 cos /3 - cos 06 sin /3 OM ON + RP RP QP ON OQ OP ~~ OP QP OP ' OQ OP .-. cos (a — /3) = cos 06 cos /3 + sin a sin /3 Next suppose that ft > 06. Then sin (06 - /3) = - sin (/3 - 06) = — sin ft cos a + cos ft sin a = sin 06 cos ft — cos 06 sin ft Also COS (06 — ft) = COS (ft — 06) = cos ft cos a + sin ft sin a .'. the theorem is true for any values of 06, ft less than 90 . Expansions of sin (a — ft) and cos (06 — ft) 61 Again sin (90 + 06 — /3) = cos (a — ft) = cos 06 cos ft + sin 06 sin ft = sin (90° + a) cos ft — cos (90 + 06) sin ft And cos (90° + 06 — /3) = — sin (06 — ft) = — sin 06 cos ft + cos 06 sin ft = cos (90 + 06) cos ft + sin (90 + a) sin ft .'. the restriction that 06 < 90 may be removed. Sim'ly „ „ „ ft „ .-. the theorem is true for all pos' A s . Lastly, if 06 = — x, and ft = — y, then sin (06 — ft) = sin {— (x — y)} = — sin (x — y) = — sin x cos y + cos x sin y = sin (— x) cos (— y) — cos (— x) sin (— y) = sin 06 cos ft — cos 06 sin ft and cos (a — ft) = cos {— (x — y)} = cos (x — y) = cos x cos y + sin x sin y = cos (— x) cos (— y) + sin (— x) sin (— y) = cos 06 cos ft + sin 06 sin ft .'. the theorem is true for all neg' A s ; and, .*. , is true for all values of 06 and ft. Note — By (JX — /3) is denoted a single angle, where 06, /3 are any angles such that /3 subtracted from OC gives its magnitude. For example : cos 13a = cos (20a — 7 Oi) = cos 20&COS 7O6 + sin 2o0£sin 7a 62 Trigonometry — Chapter IV § 20. If any one of the four results in §§ 18, 19 is assumed to be true for all values of the A s , then any other one of the four can be immediately deduced from it. Examples 1. To deduce cos (OC + ft) from sin (OC — /3) cos (a + /3) = sin {90 - (OC + (3)} = sin (9o°-a -/3) = sin (90 — OC) cos (3 — cos (90 - OC) sin j3 = cos OC cos /3 — sin OC sin (3 2. To deduce sin (OC — ft) from sin (OC +/3) sin (OC- (3) = sin {a + (-/3)} = sin a cos (— (3) + cos a sin (— /3) = sin OC cos /3 — cos OC sin /3 Exercises 1. Deduce cos (ft + (3) from sin (OC + (3) 2. Deduce sin (OC + /3) from cos (OC — (3) Any particular case, where the magnitudes of the A s are ap- proximately denned, may be proved by a geometrical construction specially adapted to it. Example To prove that cos (OC — /3) = cos OC cos (3 + sin OC sin /3 w/z«z a w about 330 , and f3 about 120 . Let a line, initially along OX, revolve in the pos' direc' thro' OC (about 330 ) into position OA ; and then back, in neg 7 direc', thro' /3 (about 120 ) into position OB. Expansions of sin (a ± /3) and cos (oc ± f3) 63 Take any p't P in OB : drop J_ s PM on XO produced, PQ on AO pro- duced, QN on XO produced, and QR on PM produced. Then OM _ ON + NM _ ON - MN _ ON - RQ _ ON ^ OQ RQ # QP OP OP" OP ~ OP OQ ' OP ~ QP " OP 0M ,„ OY Now ^5 = cos (OC — p) ON OQ = cos XOQ = cos (a - 180 ) = - cos OC ~ = cos qop = cos (180 -(3) = - cos /3 RQ OP = sin RPQ = sin QON = sin XOA = sin (360 - OC) = -sin OC ^ = sin QOP = sin (180 - /3) = sin /3 .*. cos (OC — {3) = cosOC cos (3 + sin OC sin (3 Note— In the 3rd line of the above, notice particularly the change from NM to — MN. In a case like this, great care is necessary in the substitutions for the ratios. Exercises Prove the truth of the theorems in § 18 and § 19, by special geometrical construction, in each of the following cases — 1 . OC about three-fourths of a right angle, and (3 about half a right angle : 2. OC and (3 each between one and two right angles, but OC + /3 less than three right angles : 3. OC greater than two right angles, and OC + /3 less than three right angles. 4. OC greater than a right angle, and OC + /3 less than two right angles. 5. OC sl little less than a right angle, and /3 a little greater than a right angle. 6. OC between two and two and a half right angles, and /3 between one and one and a half right angles. 64 Trigonometry — Chapter IV § 21. A number of useful developments follow immediately from the fundamental and most important theorems of §§ 18, 19. Thus, by putting 06 for f3 in the theorem of § 18, we get sin 2 a = 2 sin 06 cos 06 and cos 2 06 = cos 2 06 — sin 2 06 The latter may be put into either of the forms COS 2 06 = 2 COS 2 06—1 or cos 2 06 = 1 — 2 sin 2 06 Which again may be written COS 06 = a/(i + COS 2 0L)/2 and sin 06 = V(i — cos 2 Oijh Note — The result sin 2X = 2 sin ft cos X, is equivalent to this verbal formula — The sine of an angle is equal to twice the sine of half that angle, multiplied by the cosine of the same half angle. Hence we see that it covers all such equivalent forms as . x x sin X — 2 sin — cos — 2 2 . (X . (X X or sin — = 2 sin —; cos —? 8 16 16 or sin 16 X = 2 sin 8 X cos 8 X The last result, by repeated use of the same formula, gives sin 16 X = 16 sin X cos X cos 2 X cos 4 CX cos 8 X It should now be obvious that, by extending this to n repetitions, we should get sin 2 n a = 2 n sin X cos X cos 2 X COS4 a &c. up to cos 2 n 0( Or, in another form, x x x „ a sin a cos — cos — cos — &c up to cos — 248 c,n X 2" sin n 2 Fundamental Formulae 65 Again, by addition and subtraction, we get sin (a + /3) + sin (a - /3) = 2 sin a cos /3 sin (06 + /3) - sin (06 - /3) = 2 cos 06 sin /3 COS (06 + /3) + COS (06 — /3) = 2 COS 06 cos /3 cos (06 + /3) - cos (06 — /3) = - 2 sin 06 sin /3 A1 06 + /3 06 — /3 Also since 06 = L - L H 2 2 and ^ = ^±i_^^ 2 2 we have . a . 06 + /3 06-/3 sin 06 + sin p = 2 sin cos 2 2 . o 06 + £. 06-/3 sin 06 — sin p = 2 cos sin 2 2 Q 06 + /3 06-/3 COS 06 + cos p = 2 cos cos 2 2 o . 06 + /3 . 06 - /3 cos 06 — cos p =— 2 sin sin 2 2 Next, by division, we get sin (06 + /3) sin 06 cos /3 + cos 06 sin /3 cos (06 + /3) cos 06 cos /3 — sin 06 sin /3 sin 06 sin /3 cos 06 cos /3 sin 06 sin /3 cos 06 cos /3 , n. tan 06 + tan /3 .-. tan (06 + p) = -q v ' 1 — tan a tan p o. /1 x / /Ov "tan 06 — tan fi 66 Trigonometry— Chapter IV o i x / /Q\ cot 06 cot 3—1 So also cot (a + P) = ~ — — * COt 06 + COt ft j x / /Qv cot 06 cot ft + [ and cot (06 — ft) = ^ v ; cot a - cot /3 Whence (or from the formulae at the beginning of this §) 2 tan 06 tan 2 06 = cot 2 06 = i — tan 2 06 cot 2 06 — I 2 cot a Again, by division, we get . 06 + ft ■ a tan sin 06 4- sin p 2 sin 06 — sin ft a — ft tan — 2 Others, sim'r to the last, but of less importance, come by dividing in other ways. Two useful formulae are got thus — sin 06 sin ft sin 06 cos ft + cos 06 sin ft Since + q = ~ COS 06 COS p COS 06 COS p sin (06 + ft) .: tan 06 + tan p = * ^ COS 06 COS p o- /, P sm (^ — ft) Sim'ly tan 06 - tan ft = ^ COS 06 COS p As another development of the theorem of § 18 — sin (06 + ft + y) = sin (06 + ft) cos y + cos (06 + ft) sin y = sin 06 cos ft cos y + cos 06 sin ft cos y + cos 06 cos ft sin y — sin 06 sin ft sin y Sim'ly we should find that Fundamental Formulae 67 cos (06 + /3 + 7) = cos a cos /3 cos 7 — sin 06 sin /3 cos y — sin 06 cos /3 sin 7 — cos a sin /3 sin 7 Whence tan (06 + /3 + 7) tan 06 + tan /3 + tan 7 — tan 06 tan /3 tan 7 1 — tan a tan /3 — tan /3 tan 7 — tan 7 tan a and cot (06 + /3 + 7) cot a cot /3 cot 7 — cot 06 — cot /3 — cot 7 cot /3 cot 7 + cot 7 cot a + cot a cot /3 — 1 From the last four we have sin 3 06 = sin 06 cos 2 06 + cos 2 06 sin 06 -f- cos 2 06 sin 06 — sin 8 06 = 3 sin 06 (1 — sin 2 06) — sin 3 06 = 3 sin 06 — 4 sin 3 06 cos 3 06 = cos 3 06 — sin 2 06 cos 06 — sin 2 06 cos 06 — sin 2 a cos a = COS 3 06 — 3 COS 06 (i — COS 2 06) = 4 cos 3 06 — 3 cos a 3 tan 06 — tan 3 06 tan 3 06 = cot 3 06 = 1 — 3 tan 2 06 COt 3 06 — 3 COt 06 __ 3 COt 06 — COt 3 06 3 cot 2 a — i 1 — 3 cot 2 06 Note — These formulae can of course be got by considering 3 OL as 2 Oi + (X, expanding the function, and using the already found values for the T. F.s of 2 0i. From the formulae at the beginning of this §, we have 2 sin 2 a = 1 — cos 2 06 and 2 cos 2 06 = 1 + cos 2 06 f 2 68 Trigonometry — Chapter IV 9 i — cos 2 oc .-. tan 2 06 = I + COS 2 06 i — tan 2 a Again, since COS 2 06 = - ■ ; r - i + tan 2 oc 2 sin 06 2 sin 06 cos 06 cos 06 cos 2 06 + sin 2 oc sin 2 06 cos 2 06 2 tan 06 \ sin 2 06 = Since and sin 2 06 I + COS 2 06 I — COS 2 06 sin 2 06 sin 2 06 i + cos 2 a sin 2 06 i + tan 2 06 2 sin 06 cos 06 _ sin 06 2 COS 2 06 " COS 06 2 sin 2 06 sin 06 = tan 06 = and = cot 06 = 2 sin a cos a cos oc I — COS 2 06 sin 2 06 i + cos 2 06 i — cos 2 06 sin 2 06 From the theorems in § 18 and § 19 we have sin (06 + /3) sin (a — £) = sin 2 06 cos 2 /3 — cos 2 06 sin 2 /3 = sin 2 06 — sin 2 06 sin 2 /3 — sin 2 /3 + sin 2 a sin 2 /3 = sin 2 06 — sin 2 /3, or = cos 2 /3 — cos 2 06 The same results may also be obtained thus — sin 2 06 — sin 2 /3 = \ (1 — cos 2 06) — J (1 — cos 2 /3) = — I (cos 2 06 — cos 2 /3) = sin (06 + /3) sin (06 — /3) Useful Formulae 69 So also cos 2 /3 — cos 2 a = i (cos 2 /3 + 1) — \ (cos 2 a + 1) = • — i (cos 2 a — cos 2 /3) = sin (a + /3) sin (a - /3) Exercises 1 . Similarly prove, in each of the foregoing ways, that cos (a + /3) cos (a - /3) = cos 2 a - sin 2 /3 = cos 2 /3 - sin 2 a = cos 2 oc + cos 2 (3 — i 2. Make the following useful modifications of the preceding formulae — (1) sin 0sm

cos OC, and sin OC is neg'. /. sin OC + cos OC = — V and sin OC — cos OC = — Vi — 1 + sin 2OCI 1 — sin 2 OCJ .-. 2 sin OC = — Vi + sin 2 OC — V 1 — sin 2 OC and 2 cost OC = — vi +sin2a — v 1 — sin 2OC 1 iOC = — Vi + sin 2 OC + Vi — sin 2 OC ) 74 Trigonometry — Chapter IV 2. If 2 sin OC == — Vi + sin 20c + Vi — sin 2 a — vi + sin 2(x) + v 7 ! — sin 2 OC J then sin a + cos a = — Vi + sin 2 a and sin OC — cos a .-. cos a > sin OC, and cos OC is neg 7 . .*. a lies between — and 5 4 4 Or, in general terms, between 3 w j 5 w 2 n tt + - — and 2 n n + - — , 4 4 where n is any integer, pos', neg 7 , or zero. Exercises 1. In the formula 2 sin OC = + Vi + sin 2 OC ± Vi — sin 2 a remove the ambiguities of sign in the case when OC is between 45 and 135^ 2. In the formula 2 cos OC = ± Vi + sin 2 OC ± Vi — sin 2 a remove the ambiguities of sign in the case when OC is between 135 and 225 . 3. Determine the limits between which OC must lie, when 2 sin OC = + Vi + sin 2 OC — Vi — sin 2 OC 4. Determine the limits between which OC must lie, when 2 cos a = — Vi + sin 2 OC — V 1 — sin 2 a 5. Show that the ambiguity of sign in the formula OC . OC / ; cos — + sin — = + V 1 + sin OC 2 2 may be replaced by (— i) m , where m is the greatest integer in m „,. ,.. *. ^ .... :„ a° + 90° 360 6. Show that the ambiguity of sign in the formula OC . OC cos sin — = + Vi - sin OC 2 2 — may be replaced by (— i) m , where m is the greatest integer in -^-J- — Wolstenholme : 408. Useful Formulae J$ § 26. If A + B + C = i8o°, which is the case when A, B, C are the A s of a A, then (see Examples p. 49) .A B + C sin — = cos 2 2 , . B+C A and sin = cos — 2 2 sin A + sin B + sin C .A A .B+C B-C = 2 sin — cos f- 2 sin cos 22 2 2 A ( B + C B-C = 2 cos — - < cos 1- COS 2 f 2 2 A B C = 4 cos — COS — COS — 2 2 2 So also cos A + cos B + cos C . „ A B + C B-C = 1 — 2 sin 2 f- 2 cos cos — 2 22 . A ( B-C B + C) = 1 + 2 sin — < cos cos > . A . B . C = 1 + 4 sin — sin — sin — 222 Again, on the same supposition, cos A = — cos (B + C) and sin (B + C) = sin A .-. sin 2 A + sin 2 B + sin 2 C = 2 sin A cos A + 2 sin (B + C) cos (B — C = 2 sin A {cos (B-C) - cos (B + C)} = 4 sin A sin B sin C 76 Trigonometry— Chapter IV Using 7T for i8o°; if A + B + C = 7T, we have .A . B . C sin h sin — + sin — 2 2 2 A\ .B + C B-C cos { ) + 2 sin cos (t- 2 / 4 4 . a 7T — A . 7T — A B-C = 1 — 2 sin 2 — f- 2 sin cos 4 44 . 7T-A( B-C /7T B + C\) = i + 2 sin ] cos cos ( ) > 4 ( 4 \2 4/3 . 7T-A . 7T-B . 7T-C = i + 4 sin sin sin 4 4 4 . B + C . C+A . A+B = i + 4 sin sin — - — sin — - — 4 4 4 Note — This last result is at once deducible from the 2nd of this §, by IT A. writing — for A. 2 2 Exercises Prove that if A, B, C are the angles of a triangle, then — -_A B C IT -A 7T-B 7T-C 1. cos — + cos — + cos — = 4 cos cos COS 2 2 24 4 4 B+C C+A A+B or =4 cos cos cos 4 4 4 2. cos 2 A + cos 2 B + cos 2 C = — 4 cos A cos B cos C — 1 OA D -^ .A.B C 3. sin A + sin B — sin C = 4Sin— sin — cos — ^2 22 VI A D r, A B . C 4. cos A + cos B — cos C = 4 cos — cos — sin 1 2 2 2 5. sin 4 A + sin 4 B + sin4C = — 4 sin 2 A sin 2 Bsin 2 C 6. cos 4 A + cos 4 B + cos 4 C = 4 cos 2 A cos 2 B cos 2 C — 1 ,_ A /(sin A + sin B + sin C) (sin B + sin C — sin A) g % COS /V / — ; -— 2 ^ 4sinBsmC Useful Formulae 77 8. cos Q-^ + B - 2 c) + cos A^ + c - 2 a) + cos (— + A - 2 B ] 5 A-2B-C 5B-2C-A 5C-2A-B = 4 cos ° cos cos — 4 4 4 Math' Tri': '73. ABC 9. 8 (sin A + sin B + sin C) sin — sin — sin — v 222 = sin 2 A + sin 2 B + sin C Pet CamV: '66. Some of the preceding formulae may be very concisely expressed by means of the symbols 2, IT.* If, e. g., A, B, C are A a of a A, we have, by what goes before, A 2 sin A = 4ll cos — 2 A 2 cos A = 4 IT sin hi 2 2 sin 2 A = 4TI sin A 2 cos 2 A = — 4 II cos A — 1 A very useful additional result comes thus — Since tan (A + B) = — tan C tan A + tan B _ •'• 1 tt^l 5 = — tan C 1 — tan A tan B whence tan A + tan B + tan C = tan A tan B tan C i.e. when 2 A = 180 , 2 tan A = II tan A * The symbol II may perhaps be new to the Reader ; for it has only lately come into use in English text -books. It is the symbolic equivalent for the words the product of all such terms as. Thus, with respect to a, b, C, while 2 a means a + b + c, II a means abc ; so also II (a — b) means (a — b) (b — c) (c — a) 78 Trigonometry— Chapter IV Exercises 1. IfA+B + C = 180 , prove that, (1) 2 cot- = ncot- V J 2 ,2 (2) 2 (tan* tan ^i (3) 2 (cot A cot B) = 1 • (4) 2 cot A = n cot A + IT cosec A (e.) 2 tan — = II tan — + II sec — Koy 222 A A A (6} 2 tan — . 2 cot — = 1 + IT cosec - v ' 2 2 2 Note These formula can be deduced from the expansions of cot (OC + j8 + y), tan (OC + /3 + y), cos (a + j3 + y), sin (a + /3 + y) ^/ considering that tan 180 = o, cot 90 = o, tan 90 = 00, cot 180 = 00, cos 180 = — i, sin 90 = 1. 7/ is however advisable also to prove them independently, as in the Example given at the end of the preceding page. 2. In the manner indicated in the above Note ; and with the notation of § 22, prove that (m being any integer) i°, if 2 OC = m it, then t x + t 5 + t 9 + &c = t 3 + t 7 + t u + &c ; 7T and, 2 , if 2 OC = (2 m + 1) — , then 1 + t± + t 8 + &c = t 2 + t 6 + t 10 + &c. 3. IfABCDEFisa hexagon, prove that 2 (tan A tan B tan C) = 2 tan A + 2 (tan A tan B tan C tan D tan E) How many terms will there be in each 2 ? E. T. xl. 4. If +

) 1 — sin 6 1 — sin ) (1 - sin 6) (1 — sin ) Examples 8$ i. e. if (i - sin 2 0) (i - sin c/>) 2 - (i - sin 2 cp) (i - sin 0) 2 = 2 (sin — sin cp) (i — sin 0) (i — sin c/>) i. e. if (i + sin 0) (i — sin ) (i - sin 0) = 2 (sin — sin cp) which is the case. E. T. xlviii. 77 6. ^x = — j /ra^ Matf cos 3 X — COS 2 X + cos x = A. i?. U. Matric' : '8j. If we put X for cos 3 x — cos 2 x + cos x then X = cos 3 x + cos 5 x + cos x, since 2 x + 5 x = 77 .-. 2 X 2 = 3 + cos 10 x + cos 6 x + cos 2 x + 2 cos 8 x + 2 cos 2 x + 2 cos 6 x + 2 cos 4 x + 2 cos 4X + 2 cos 2 x — 3 — 5 cos 3 x — 5 cos x + 5 cos 2 x = 3-5X .-. ( 2 X-i)(X + 3) = o But X =£ - 3 .. X - ^ 7. If k> B, C are angles of a triangle, and if sin 3 co = sin (A — co) sin (B — co) sin (C - co) prove that cot co = cot A + cot B + cot C ; and deduce that cosec 2 co = cosec 2 A + cosec 2 B + cosec 2 C. We have II cosec A = II j- 8 !" ( A A 7 ^ l (sin Asm co J = II (cot co — cot A) .-. cot 3 co — cot 2 co S cot A + cot co 2 (cot A cot B) — II cot A — II (cosec A) = o But (see p. 78, Ex's 3, 4) II cot A + II cosec A = 2 cot A and 2 (cot A cot B) = 1 .-. cot 3 co + cot co = (1 + cot 2 co) 2 cot A .*. rejecting the factor 1 + cot 2 co, which would give only imaginary values for cot co, we get cot co = Scot A G 2 u Trigonometry— Chapter IV Then squaring, and recollecting that 2 2 (cot A cot B) = 2, we get 1 + cot 2 o) = 2 cot 2 A + 3 .-. cosec 2 o) = Scosec 2 A Note — The Student should carefully notice these results. 60 is called the Brocard Angle* of the A whose /\ s are A, B, C; and has numerous important properties : some of these will be given in Chapter XIV. \ °=J\ , 1 + e u 8. Given that tan - = V • tan - 2 1 — e 2 and r = a (1 - e 2 ) 1 + e cos /- . /— 7 ; . U to prove that Vr .sin - = V a (1 + e) . sin - A and Vr . cos - = Va (1 — e) . cos - Christ's CamV: '47. The 2nd of the methods given in Example 2, p. 30 will be useful here. Thus assume that , . 6 , . u \ A sin - = v 1 + e sin - - where A has to be found. and A cos — = V* — e cos - 2 2 Then, squaring and adding, we get A 2 = (1 + e) sin 2 - + (1 — e) cos 2 - = 1 — e cos u v J 2 2 whence A = a/i — e cos u Again since tan 2 — = tan 2 — 2 1 — e 2 1 — cos 1 + e 1 — cos u 1 + cos $ 1 — e ' 1 + cos u _ (1 — e) (1 + cos u) - (1 + e) (1 — cos u) (1 — e) (1 + cos u) + (1 + e) (1 — cos u) cos u — e 1 — e cos u * See Eticlid Revised, pp. 394-402. Exercises 85 .*. 1 + e cos0 = 1 - e* 1 — ecos u \ 1 — e cos u = 1 - e* 1 + e cos a .*. A. = vr/a Whence the req'd results are seen to be true. 9. ABCD is a line trisected by the points B, C : P is any point on the circle whose diameter mBC : if the angles APB, CPD are respectively denoted by 0, (j), it is required to show that tan tan

(2 2 2 2) _ a + 8 (39) 2 sin a - sin (OC + j3 + y) — 4 II sin oc + 8 (40) 2 cos a + cos (a + /3 + y) =r 4 n cos - _ sin (OC + 8 + y) . _ (41) 2 tan OC ^ *— = II tan OC cos OC cos p cos y v ^ cos (ft + /3 + y) „ (42) 2 cot oc + - — . v, — r^- e n cot a sin OC sin p sin y Note — 7%g Student should compare these last four identities with the par- ticular conditional cases of them given on pp. 77> 7^« (43) 2 sin 2 a sin (/3 - y) E 2 sin (OC + (3) x 2 sin (/3 - OC) (44) 2 cos 2 a sin (/3 - y) E 2 cos {OC + j3) x 2 sin (/3 - OC) Wolstenholme : 467. Exercises 89 4. If sin /3 = m sin (2 a + /3), prove that os 1 + m x tan (a + B) = tan OC K ^ 1 — m Z XV 5. If tan - = tan — tan — , prove that 2 2 2 tan z = 2 sin x sin y cos x + cos y 6. If sin x sin y = sin (OC + (3) sin y ' cosx cosy = cos (0C + (3) cosy and cos 2 x + cos 2 y = 1 + cos 2 (OC + (3 + y) prove that sin 2 (OC + /3) + sin 2 y = sin 2 (a + /3 + y) 7. If tan = m, and tan qf) = h, prove that a . 2 (m + n) (1 - mn) sin 2 (0 + d>) = — =— i sr- v ^ (1 + m 2 ) (1 + n 2 ) . cos oc — cos (3 8. If COS 6 = - n , prove that 1 — cos OC cos p /1 tan ~ 6 2 tan — = + tan l- 2 9. If sin 2 OC + sin 2 /3 - 2 sin a sin (3 Cos (OC - (3) = n 2 (sin a - sin /3) 2 (X 1 + n ± 3 prove that tan — = — = — cot — r 2 1 + n 2 10. If vers OC = x, vers/3 = mx, vers y = 1 — m, and OC + (3 = y, show that x = 1 + \J ( ) 11. If sin x cos y = tan OC cot y sin y cos x = tan /3 cot y and cos 2 y — cos 2 x = cos 2 y prove that sec 2 OC — sec 2 /3 = sin 2 y 12. If tan - = tan 3 — , and tan Q = 2 tan d>, show that is 2 2 r- • 2 one value of ) =V(b 2 - a 2 ) (a' 2 - b ,2 )/ab' + a' b and that cos (0 ± c/>) = aa' + bb' /ab' + a'b i n> xr * a xsina , ysinft 17. If tan = , and tan cp = — -, y — x cos OC ^ x — y cos OC show that tan (0 + c/)) = — tan OC 18. If tan 2 = tan {0 - OC) tan {0 - (3) 2 sin OC sin /3 prove that tan 2 = sin (a + (3) 19. If a cos + b sin = c ) and a cos c/) + b sin c/> = c j prove that tan {0 + c/>) = 2 abAa 2 - b 2 ) _ -. Tr , sin a sin 20. If tan c/> = ^. t7. SchoV: '87. /*// Caw£': '87. cos - cos a sin OC sin c/> prove that tan = cos (j) ± cos CX il/tfM' Tri'\ '68. 21. If tan = and tan (j) = Exercises 9 1 sin OC cos y — sin (3 sin y \ cos CX cos y — cos (3 sin y sin a sin y — sin /3 cos y cos OC sin y — cos /3 cos y find tan (0 + (fr) in its simplest form cos OC 22. If sin a = , find tana Vi — m 2 sin 2 OC 23. If tan 466 15' 38" = — ^, find the sine and cosine of 2 3 3 7' 49" / (3) tan + tan (p (5) cos 2 + cos 2 (2) cos cos (j) (4) tan - + tan -*- (6) cos 3 + cos 3 (J> 2 2 Wolstenholme : 462. 9 2 Trigonometry - Chapter IV 29. If A + B + C = 360 , prove that A D 2 sin A (1 + 2 cos B) = — 4 II sin 2 Afo^' 7W': '64. 30. If V2 cos A = cos B + cos 3 B os 3 B I si n 3 B ) and V2 sin A = sin B — sii prove that + sin (B — A) = cos 2 B = -i Math' Tri': '71. 31. If x sin co = X sin (co - a) + Y sin (a> - (3) ) and y sin co = X sin OC + Y sin/3 ) show that x 2 + y 2 + 2 xy cos co = X 2 + Y 2 + 2 XY Cos (OC - j3) Note — The given eq*ns are those connecting the coord's (x, y) of a p't, referred to axes Ox, Oy, inclined at co, with its coord' s (X, Y) referred to axes OX, OY, making A s OC, /3 with Ox. Cf. Salmon* s Conies^ pp. 7, 9. 77z£ result shows that the expression x 2 + y 2 + 2 xy cos x Oy is invariable, no matter how the axes are turned about O. 32. If OC, /3, y, b are the angles of a quadrilateral; prove that / n v ■ .. cx + /3./3 + y.y + a (1) 2i sin OC = 4 sin sin L sin ' 222 /N „ 0C+ ft 8 + y y + 0C (2) ZcosOC = 4 cos — cos cos ' 222 33. If OC, {3, y, 6 are the angles of a quadrilateral ; and if 6, (j) are the angles formed by producing each pair of opposite sides to meet ; prove that ay j8 . a d

1. From this cos 18 = Vi — sin 2 18 = \ *Jio f 2 V$ 104 Trigonometry — Chapter V Then tan i8° = ^S- 1 - A / 6 - 2 a/ 5 *JlO+2 V& V 2 >/5 (^5 + J ) (3 ~ A) (^5 ~i)^ | 20 J Since 72°= 90 — 18 , the T. F.s of 72 are at once deducible from those of 18 . Next cos 3 6°=i-2sin^8°=i- 6 - Q 2 ^ = + ^5+1 ■. sin 36° = Vi — cos 2 36° = \ v/io —2 a/ 5 a + o« „*° x/io-2^5 _ A / 2 ^5 (-/5-1) (3- ^5) then tan 36 = ^ = A/ — -=- -7=— v 5 + 1 V 2 (3 + Vs) (3 - v 5) 4 Since 54 = 90 — 36 , the T. F.s of 54 are deducible. And, since 3 = 18 — 15 , the T. F.s of 3 can be found. Again sin 9 = \ {Vi + sin 18 — Vi — sin 18 } = i {\/3 + V5 - x/o - A} Sim'ly cos 9° = i is/s + V5 + xA - A } It now becomes clear that the T. F.s of numbers of other A s , such as 27 , 42 , 63°, 8i°, 87 , &c can be deduced from the fore- going. Exercises 1. Prove that — (1) sin 2 24 - sin 2 6°= 5 ~ * o (2) tan 54 = Vi + f V5 T. F.s of some special Angles 105 (4) tan 9 = VI + 1 - V 5 + 2V5 (5) sin 3 = 1 {(^5 -1) V2 + Vl- V{io + 2 V5) (2 - 71)} (6) cos 12 = |- { V5 - 1 + -/30 + 6 V5} (7) cos 27 = J { ^5 + V5 + Vs - VI} (8) tan 27 = VI — 1 - ^5 - 2 V5 / n . ^ o Vio + 2 Vk + V\ — 1 (9) sin 6? = ° ■ " 5 4V2 (10) sin 87 = | {(V5 - 1) aA - V3 + V(io + 2 V5) (2 •+ VI)} t \ x IT . 37T I (11) tan — tan — = — 10 10 Vg 2. Find cos 42 to five decimal places. 3. If tan = -, and tan /I.s!n^, V^.sinJ and- n/I . sin ^ 2 IO 2 6 2 IO Wolstenholme : 405. 106 Trigonometry— Chapter V § 34-. In the last paragraph we found, by starting with the eq'n sin 2 06 = cos 3 06, that sin i8° is a root of the eq'n 4 X 2 + 2 X — i=o If we had started with the equation sin 3 06 = COS 2 06, we should have found sim'ly that sin 18 is a root of the eq'n 4X 3 — 2X 2 — 3X+ 1=0 The process employed may be generalized. For, when n is even, the expressions sin n 06/cos 06 and cos n 06 ; and when n is odd, the expressions sin n 06 and cos n 06/cos 06 are each expressible in terms of powers of sin 06 only. This has been seen to be the case for small values of n ; and will hereafter (in Chapter XVIII) be proved to be true in all cases. 7T Now if m 06 + n 06 = — , where m + n is an odd integer, 2 one of the two m and n must be odd. .-. , since we have the two eq'ns sin m 06 = cos n 06 and cos m 06 = sin n 06 each of these is reducible to terms involving powers of sin 06 only. .-. each of them will give an eq'n in x of which sin 06 is a root. So also if m06+n06 = 7T then sin m 06 = sin n 06 and cos m 06 = — cos n 06 Now, when m, n are both odd, sin m 06, sin n 06 are each expressible in terms of sin 06. And when m, n are both even, cos m 06, cos n 06 are each expressible in terms of sin 06. Examples and Exercises 107 .*. in either of these cases we should get an eq'n in x of which sin OL is a root. Sim'ly if m 06 + n OC = 2 ir, &c. Examples If OL = io°, then 5 OC = 50 and 4 OL = 40 .\ sin 4OL = cos 5a .-. 4 sin OC cos OC cos 2 an( l so on « Example and Exercise 109 fM*)=o J7/6J= CO The successive changes off (6) , as the generator of passes thro' the four Tivcf j(0) = quadrants, are shown in the accompanying fig / . f(3Tll2)=0 The graph of f(6) is given below ; where OX, OY are rect'r axes ; and Aj, A 2 , &c B l5 B 2 , &c are taken so that 'A* x OA x — A x A 2 = A 2 A 3 = A 3 A 4 , and each represents a r't A- OB x = JAjA, = B 2 A 2 = A 2 B 3 = B 4 A 4 B x Aj = -q A x A 2 = A x B 2 = B 3 A 3 = A 3 B 4 The ordinates of p'ts on the curve, corresponding to the abscissae for /\ S 0C, j3, give the relative values off{0C),f((3). Exercises Trace the changes in sign and magnitude of /(d), as $ changes from o to 2 77, and draw its Graph, in each of the following cases — 1, /(#) = sin 6 - cos0 ' Previous. ,' CamV ': '79. 110 Trigonometry — Chapter V 2. f(0) = sin (77 cos 0) 3. /(0) - tan I * (sin + cos 0)1 4. /(0) = sin + sin 2 5. /(0) = 0sin0 6. /(0)=^ v cos 7. /(0) = sin 0. sin 2 8. /(0) = tan0 + tan 2 9. /(0) = sin0 + cos 2 -in /vfl> 2 sin 0- sin 2 10. /W- 2 S i n ^ + sin 2 . - .0 sin + 2 sin — 11. /(0)=- 5- sin — 2 sin — 2 § 36. By using the numerical values of T. F.s of special A s , we may — i°, modify some of the formulae already proved; and, 2 , prove important new formulae. Examples 1. Since tan 45 = 1, we have , _ 1 + tan OC tan (45 ± a) - ^=r-. — - K 1 + tan OC From this we get that tan (45 + a) - tan (45 - a) 1 + tan OC 1 — tan OC 1 — tan OC 1 + tan OC 4 tan OC 1 — tan 2 OC = 2 tan 2 OC Examples in 2. Since sin (36 + OC) - sin (36 - a) = 2 cos 36°sin OC + V5 + 1 sin a and sin (72 + a) - sin (72 - a) = 2 COS 72 sin a + Vs - 1 . = - sin OC 2 .*., subtracting and rearranging, we get sin OC + sin (72 + a) — sin (72 — oc) = sin (36 + OC) — sin (36 - oc) (Siller's formula) Again cos (36 + a) + cos (36 — OC) = 2 cos 36 cos a + V5 + 1 and cos (72 + oc) + Cos (72 — OC) = 2 cos 72 cos OC + V5 — 1 cos a cos a Whence ^ 2 cos a + cos (72 + a) + cos (7 2 - OC) = cos (36 + a) + cos (36 - OC) Note — These two results are called ferrhulce of verification \ because they are specially useful as a means of verifying the accuracy of a table of numerical values of the sines and cosines of A s - The verification will be effected by giving any value to OC, and then substituting, from the table, the numerical values of the corresponding sines and cosines. Notice that either formula is deducible from the other by writing for OC its complement. 3. Two circles touch each other, and each touches each arm of an angle : ifR. is the radius of the larger, and t of the smaller, prove that R = r tan' (ft + IT A Let CSD be . * cos - — sin — _4 V i — tan J R = r tan 2 ° + ^ = J / J ~ , \ — 2 n/ V 2 — >/ 2 — V 2 — &c — V 2 where V occurs n times ; and the + or — sign is taken as n is even or odd. I ii4 Trigonometry — Chapter V 45° / / 26. 2 cos — = V2 + V2 +&cton terms 27. sin 2° sin 14 sin 22 sin 26 sin 34 sin 38 sin 46 x sin 58 sin 62 sin 74 sin 82 sin 86° = 2- 1 8 1 - tan 3 OC . 28. If tan - = 7 — 3-^, then 2 1 + tan 3 OC acot2a = co t *£-f)-tan*(H) sin a cos /3 ., 2£>If tan * = 53^TT^' then *n£-*»?*»(*-?).«--o<*foa.(*-f) Note— /W x for tan ^ , 0^ jo/w M* quadratic J 2 2X /(i - x 2 ) = sin a cos /3/ (cos OC + sin /3) __ Ssin OC - 1 Tf 30. -^ — = II 2 cos a A- tan V I+tan vJ where the symbols apply to OC, (3, y ; and OC + /3 + y = 77/2 Afa/// TV?': '66. 31 + a / - ± - V ! — sin 2 (— + 2 CX j equally represents sin (- + a) and cos (— + OLJ"; and remove the ambiguity of signs when OC is between 77/4 and 77/2 77 277 377 477 5 77 677 7^7 /l\ 7 32. cos - cos - cos — cos — cos — cos — cos — = y Math' Tri' : f 66. 33. (sec 2 - + sec 2 — + sec 2 ^) x (cosec 2 - + cosec 2 — + cosec 2 — — ) = 192 \ 7 7 7 / Ood Jun f Math' Schol' \ '87. Exercises n5 34 sin = 5 sin ^ . if tan Q + ?)- tans (^ + £) , then ( i + sin 2 $ cot 2 — j f i + sin 2 , as to magnitude -by the species of the auxil' A; and, 2°, as to sign „ X and base „ „ „ ; .-., in order that the tangents of two A s may be identically equal, their corresponding auxil' A s must — i°, be sim'r so that their J_ s are homologous; and, 2°, have the X and base of either A of the same or dif- ferent signs, according as the X and base of the other A are of the same or different signs. MX X' M O N £ or, if neg', < — f the expression under the radical will be neg 7 , and .-. the value of tan imaginary /. tan 6 = \ \ 2 n + 1 + >/ 4 n 2 + 4 n - 15 } where n may be any integer, excepting 1, o, — 1, — 2. Exercises 1. Show that, if cos 6 = o, then all values of 6 are included in m 77 + 77/2 ; where m is any integer, zero included. 2. Show that, if cos (X = — 1, then all values of OC are included in m . 1 8o° ; where m is any odd number. 3. If tan CX = 1, show that OC — (4 n + 1) 45 . 4. Find all values of 6 which satisfy the equation tan 2 = tan 2 77/4. 5. If cos 6 = cos (X, and tan 6 = tan OC, simultaneously (where a is a known angle) find an expression containing all values of 0. 6. If sec = sec OC, and cosec = cosec OC, simultaneously (where OC is a known angle) find an expression containing all values of 0. 7. Are there any angles which have both the same cosine and the same cosecant as a known angle OC ? Exercises I2 7 8. Find all the values of satisfying each of the following equations — (i) sin 2 = fcos0 (2) sec 2 - fsec0 + 1 = o (3) 6 cot 2 = 1+4 cos 2 (4) sin 3 + cos 3 = — V 2 (5) sin + sin 2 + sin 3 = o (6) cos n + cos (n - 2) = cos (7) 2 cot 2 - tan 2 = 3 cot 3 a Q ($) 8 cot = sec 2 - + cosec 2 - v J 2 2 (9) sec 2 + cosec 2 = 3 sec 4 (10) cosec 3 + cosec 2 = sin cosec 2 cosec 3 (11) sin (0 + a) = cos (0 - a) (12) sin (0 + a) + cos (0 + a) = sin (0 - Oi) + cos (0 - a) (13) sin a + sin (0 - OC) + sin (2 + a) = sin (0 + a) + sin (2 - a) - (14) cosec 2 sec 2 - = 2V3 cosec 2 (15) sin0 - cos0 = 4 cos 2 sin (16) cos 3 = sin (17) sin 3 = cos ; and find which of its solutions satisfies 1 + sin 2 = 3 sin cos (18) 3cos 2 + 2 VI cos0 = 5i Woolwich : '81. (19) sec 4 — sec 2 = 2 Math' Tri'\ '77, (20) cosec 4 a - cosec 4 0= cot 4 a - cot 4 Math! Tri': fl j6. (21) (1 + sin 0) (1 - 2 sin 0) 2 = (1 - cos a) (1 + 2 cos OC) 2 Math' Tri': '62. i28 Trigonometry — Chapter VI , N i i sin 2 ( 22 ) T^TTn + sin30 sin 2 sin0sin30 (23) cot — tan = cos + sin (24) tan ( — — sin#J = cot / — -^cos0j \2 v 2 r \2 ▼ 2 * Caius CamV \ '82. (25) 7 COS.3 = sin 2 + cos 2 C7ar£ fl»fl? Caius CamV: '74. (26) 4 (1 + 8 cosec 2 40) = (tan 2 + cot 2 0) (tan 2 a + cot 2 a) (27) tan 5 = 5 tan 9. If vers = vers (0 — OC), where (X is a known angle (not a multiple of 7r) find 0. 10. Show that all the solutions of tan + cot = 4 are contained in (n + -j- + -g-) 77 1 1 . Show that all the solutions of sin 2 2 - sin 2 = \ are contained in (n + ^ + ~q) it 12. If n lies between certain values, the equation tan 3 = 1\ tan can only be satisfied by = m tt : find those values. Solve the equation when n = — 1. Show that if = 7r/i2, then n = 2 + V3 13. If tan (cot 0) = cot (tan 0) show that sin 2 = 4/(2 n + 1) 7T where n is any integer, excepting — 1, or o. 14. Show that the equation sin + sin ) can only be true when one of the quantities 0, (p, + (f) is a multiple of 2 7T. Johris CamU \ '33. 15. Five angles are in arithmetical progression ; and the cosine of the mid angle is equal to the sum of all their cosines : find the difference of the angles. Exercises 129 16. If A is the greatest angle of a triangle and 2 sec A cosec A — cosec A cot A = sec A, find how many degrees there are in A. /. /. : '88. 17. Find the six positive angles, less than four right angles, satisfying 8 sin 3 9 — 6 sin + 1 =0. Johris Camb' : '83. 1 8. Show that all the values of satisfying sin 3 = sin cos 2 9 are contained in nir/2, where n is any integer. 19. If sin 9 + sin 22. If sin 9 + sin 2 9 + sin 3 = sin OC and cos 9 + cos 2 + cos $9 = cos (X prove that (X is a multiple of 77. Note — Square and add. 23. If b 2 = 4 ac, x 2 + y 2 = (a + c) 2 Caius Camb': 78'. and x = a cos 9 + b cos 2 + c cos 3 9 y = a sin 9 + b sin 2 v + c sin 3 os 2 9 + c cos 3 0} in 2 9 + c sin 3 9) prove that = (2 n + 1) 77/2 ; and that there are no other solutions. Pet' Camb f -. '66. 24. If cos 2 m0 {tan(m + n) + 1} = cos 2 n0 {tan (m + n) 9 -1} show that all values of are contained in the two forms r 77/2 (m + n) and (4 r + 1) 77/4 (m — n) 77, where r may have any integral value, including zero. K 130 Trigonometry — Chapter VI 25. Solve each of the equations — (1) l6 cos2x + 2sin2x + 4 2Cos2 X = 4Q (2) 73 4 6x7 secx + 7 I+se cx_ 7oIO x 7 2secx _ 7 3+ 2 secx +3x72 + 3 secx ==I47# Bordage : E. T. xlviii. 26. If sin OC + sin /3 + sin y = o J and cos a + cos/3 + cosy = 0) prove that 2 sin 2 OC = 3/2 = 2 cos 2 (X. Wolstenholme : 435. Note — ,£W sin y, COS y 0?? 02"/z£r sides: divide corresponding sides, giving OC + (3 = 2 m 77 + 2 y .* two sim'r results follow. Then square given ecfns and subtract. Restdt now follows easily. § 4-1. The preceding general results may also be used to ex- plain some ambiguities which have occurred in formulae already proved, or which may occur in similar formulae; and indeed to show a priori that such ambiguities are to be expected. The general principle of the explanation is this — If f{9) represents a T. F. of 6 ; and if we are given the value of _/*(#) to be a known quantity (k say) we are not given 0, but only that it is one of the A s contained in the general expression corresponding to f(0). Hence, in finding another T. F., say F (xr\ 0), where m is integral or fractional, we have an infinite number of A s which, though they give the same value for f (0), do not necessarily give the same value for i^(m 6). To get the values of F (m 6) corresponding to the A s which satisfy f(6)=k, we must substitute from the general value of 0, satisfyingy(#)=k, in F(m 6). When this is done we shall arrive at an eq r n con- necting f{0) and ^(m 6), from which for a given value of the one we can find the corresponding values of the other. We can also apply to such an eq'n the algebraic connection between its roots and coefficients, and thus deduce formulae which might other- wise be more troublesome to arrive at. The discussion of some cases will make all this clearer. Examples 131 Examples 1. To explain the ambiguity in the formula sin 6 = ± Vi — cos 2 where it is implied that cos is given, and si n wanted. If we are given the numerical value (k say) of cos 0, we are not given 0, but only that it is one of the A s included in 2 n it ± where is any one definite A satisfying the eq'n cos = k. Now, in finding sin from cos 0, we have an infinite number of A s > which, though they give the same value for the cosine, do not necessarily give the same value for the sine. .*., in the formula Vi — cos 2 = sin 0, we may expect a priori to get not only sin but also the sines of all A s included in 2 n it ± 9. But sin (2 n it ± 6) = ± sin 6 i.e. we may expect h priori to get two values for sin 0, equal in magni- tude, but opposite in sign, corresponding to any given value of cos 6 ; and hence the ambiguity in sin0=+Vi — cos 2 6 is explained and justified. 2. To explain the ambiguity in the formula n /l + COS 2 cos 6= + v — 2 where it is implied that cos 2 6 is given, and cos is wanted. If we are given the numerical value (k say) of cos 2 0, we are not given 2 but only that it is one of the A s included in 2 n 77 + 2 where 2 is any one definite A satisfying the eq'n cos 2 — k. .'. , when cos is expressed in terms of k, all its values are included in cos (n 77 + 0) Now, if n is even this is cos ; but, if n is odd it is — cos 0. Hence, a priori we see that cos when expressed in terms of cos 2 must have two values, equal in magnitude, but opposite in sign ; and so the ambi- guity of sign in the formula cos = + \J ( ) is explained and justified. K 2 [32 Trigonometry — Chapter VI 3. To show how it is that four values arise out of the expression ■| {\/i + sin 2 + Vi — sin 2 0} If we are given the value (k say) of sin 2 0, we are not given 2 0, but only that it is one of the A s included in n 7T + (- i) n 2 where 2 is any one particular A satisfying the eq'n sin 2 = k .-., when sin is expressed in terms of k, all its values must be contained in sin {mr/2 + (- i) n 0} Now, i°, if n is even (say 2 m) then this is sin (m 77 + 0) Expanding; and recollecting that sin m 7T = o, and cos m TT = + 1, we get + sin 0. Also, 2 , if n is odd (say 2 m + 1), then above becomes sin (m tt + 17/2 — 0) Expanding; and recollecting that sin m tt = o, cos m 7T = ± 1, and sin (77/2 — 0) = cos ; we get + cos 6. Hence we see that the^^r values of the expression {Vi + sin 26 + Vi — sin 2 d}/2 which we found (in § 25) to be + sin and + cos 0, may be expected a priori. 4. To see what can be deduced from the formula 4 cos 3 — 3 cos = cos 3 If cos 3 has a given value (k say) and if x is put for cos 0, then the formula gives 4 x 3 — 3 x — k = o as an eq'n to determine x in terms of k. About this eq'n we know, by theory of eq'ns, that i°, it has three roots ; and that if these are denoted by x x X2 x 3 , then 2 , x x + x 2 + x 3 = o 3 J X l X 2 + X 2 X 3 + X 3 X l = "~ "4 4°, x x x 2 x 3 = k/4 Whence we see that cos expressed in terms of cos 3 should have three values ; and then that three eq'ns connect these values. Now we are not given 3 but only that it is one of the A s contained in 2 n tt ± 3 Examples 133 where 3 is any one particular A satisfying cos 3 = k .'. all values of x are contained in cos (2 n 77/3 + 0) If n = 3 m , this is cos (2 m 77 ± 0), i. e. cos ,> 3f"-i, „ cos (2 m 77 - 2 77/3 ± 0), i. e. cos (2 77/3 ± 0) „ 3 m + I, „ COS (2 m 77+ 277/3 ± 0), „ COS (277/3 ± 0) But n must be one of these forms. .\ the values of x are cos 0, cos (2 77/3 + 6), cos (2 77/3 — 0) Whence it follows that 4X 3 - 3 x - cos 30 — (x - cos 0) (x - cos 2 77/3 + 0) (x - cos 2 77/3 - 6) Also (as may easily be proved independently) cos 6 + cos (2 77/3 + 6) + cos (2 77/3 — 6) = o cos 6 cos (2 77/3 + 6) + cos (2 77/3 + 6) cos (2 77/3 — 6) + cos (2 77/3 - 0) cos 6 = -f cos cos (2 77/3 + 0) cos (2 77/3 - 0) = (cos 3 0)/4 Again 2 x x 3 = (S Xj) 3 - 3 2 x x 2 (x 2 + x 3 ) - 6 x x x 2 x 3 = 3 2 Xj 3 - f cos 3 .-. 2x! 3 = f cos 3 i. e. cos 3 + cos 3 (2 77/3 + 0) + cos 3 (2 77/3 — 0) = |. cos 3 Again 2 x x 2 = (2 xO 2 - 2 2 x x x 2 = f 2x 1 4 = (2x 1 2 ) 2 - 2 2x 1 2 x 2 2 = (| - 2 {(2x x x 2 ) 2 - 4x x x 2 x 3 2 xj £ 9 _ 9 4 F ~~ 8 i. e. cos 4 + cos 4 (2 77/3 + 0) + cos 4 (2 77/3 — 0) = § Note — This last result has been given (see p. 113) to be proved independently. It is clear that the formula 3 sin — 4 sin 3 = sin 3 and all analogous formulae may be treated sim'ly. 134 Trigonometry — Chapter VI 5. To give another Example. n a, tan — c, tan 3 6 + &c Since tan n = — - — w-~ : — tj: =— 1 - c 2 tan 2 6 + c 4 tan 4 - &c .*. , if we put x for tan 0, and suppose (e. g.) n 6 = it, then c x x — c 3 x 3 -t- c 5 x 5 — &c ± c n x n = o is an eq'n whose roots are the values got by letting m have, in succession, the values o, 1, 2, &c . . . n — 1 in tan (m + 1) or/n So also we may treat the formula a cot n - c 2 cot n " 2 + c 4 cot 11 " 4 - &c cot n = ■ l - — ^ c 1 cot n - 1 ^-c 3 cot n - 3 + c 5 cot n - 5 0-&c Exercises 1. Show a. priori that when tan OC is expressed in terms of sin (X, two values should result. 2. Determine a priori the values tan OC has when expressed in terms of tan 3#. Form the equation whose roots are those values ; and deduce that tan OC tan (6o° + OC) tan (6o° - a) = tan 3 OC and tan OC + tan (6o° + OC) - tan (6o° - OC) = 3 tan 3 OC 3. Show that if sin OC is determined from a given value of sin 3 OC it should have three values ; and that the sum of these is zero. 4. Given the value of sin OC show a priori that sin %0C will have eight different values, when expressed in terms of sin OC. 5. From the formula 3 cot OC - cot 3 OC COt 3 OC = 75 1-3 cot 2 OC deduce three formulae analogous to those found in Example 4. 6. If sin OCA is found from a given value of cot OC, find how many values of sin OC/3 may be expected d priori. 7. Prove d priori that sin n OC, when expressed by sin OC, will have one or two values, according as n is odd or even ; and that cos n OC, expressed by cos OC, will have only one value : n is a positive integer. Exercises 135 8. Show that tan OL, expressed in terms of sin 4.OL, will have four values ; and find these values. John's Cam}/ ; '40. 9. If sin OL is expressed as a function of sin 6 OL, determine by a priori reasoning how many values the resulting expression ought to have for a given value of sin 6 OL. Pet' CamV : '61. 10. Form the equation for determining tan OL in terms of tan 6 0L; and show a priori that it ought to have six values. Prove that the sum of these six values is — 6 cot 6 OL. Pet' Camb': '60. 11. Prove that tan + tan (17/5 + 0) + tan (2 77/5 + 0) + tan (3 77/5 + 0) + tan (4 77/5 + 0) = 5 tan 5 0. Q. C. B. '88. 12. Prove that x 3 — 3 x 2 tan 3 6 — 3 x + tan 3 6 = (x - tan 0) (x - tan TrA + 0) (x - tan 2 77/3 + 0) R. U. Schol': '83. 13. Prove that cosec 9 + cosec (77/3 + 6) + cosec (77/3 — 6) = 3 cosec 3 Christ's CamV\ '78. 14;. If sin 3 is given, and thence tan is found, show a priori that six values are generally to be expected. Hence show that tan 2 tan 2 (77/3 + 6) + tan 2 (77/3 + <9)tan 2 (77/3 - 6) + tan 2 (77/3 - 0) tan 2 = 6 sec 2 3 + 3 Clare, Caius and King's Camb r : '8o. 15. Form the equation whose roots are tan 2 77/11, tan 2 277/11, tan 2 377/n, tan 2 477/n, tan 2 577/11 Christ's, Emmf and Sid' Cami/i '82. 16. Prove that (x — 2 cos 2 77/7) (x — 2 cos 477/7) (x — 6 77/7) Ex 3 + x 2 — 2 x — 1 136 Trigonometry — Chapter VI 17. Prove that — cos 11 77/36 + cos 13 77/36 -f cos 35 77/36 = o cos 11 77/36 cos 13 77/36 + cos 13 77/36 cos 35 77/36 + cos 35 77/36 cos 1 1 77/36 = — 36 cos 1377/36 cos 3577/36 = ^— — 18. Prove that ill cos 2 77/7 + cos 2 ^> cos 4 77/7 + cos 2 <£ cos 6 77/7 + COS 2 (p _ 7 tan 7 (f) — tan ^ 2 sin 2 d> i?. £7. Schol': '86. Note — Express {see Example 5) tan 7 hi terms of tan (j) : the result will give an eq'n of the *]th degree in tan (p, the sum of whose roots is 7 tan 7 (f). These 7 roots will be tan (f), tan (77/7 ± (p), tan (2 77/7 ± $), tan (3 77/7 + $) ; and these added will give the result. 19. Assuming that the sum of the cubes of the roots of the equation x 11 + PiX 11-1 + PaX 11 " 2 + &c + Pn.iX + p n = is - Pi 3 + 3 Pi P2 - 3 Ps prove that cot 3 + cot 3 {6 + 77/n) + cot 3 (6 + 2 77/n) &c, to n terms, = n cot n 6 (n 2 cosec 2 n — 1) R. U. Schol': '90. Note — If x = cot 6, then {Example 5) the expansion of cot n 6 in terms of "x gives x n -ncotn^.x n - 1 + n(n ~ l) x n - 2 2 n (n - 1) (n - 2) . n_ 3 + — - ~^ cot n 6 . x^ + &c = o, o . whence result easily follows. 20. Prove that cot 2 77/11 + cot 2 277/n + cot 2 377/11 + cot 2 4 77/1 1 + cot 2 5 77/1 1 = 15 Ox' Jun' Math' Schol': '90. CHAPTER VII On Angles expressed as the Inverses of Trigonometrical Functions § 42. If sin X = 06, the fact implied by this direct statement could be otherwise expressed by, what may be termed, the inverse statement — X is the angle whose sine is 06 But this latter statement is ambiguous ; for we know that there is an infinite number of A s whose sines are each 06. Hence, if we are to use such ' inverse ' statements, we must have some convention to limit their ambiguity. The convention usually made is this — When an angle is expressed as the inverse of a T. F. the least positive angle for which the function has the specified value is always to be understood, unless the contrary is expressly stated. Now inasmuch as the phrase — ' the least positive angle whose sine is Oi ' — is too tedious for general use, some brief notation to express these words is desirable. The notation usually adopted is * sin -1 06 ; with a similar notation for A s expressed inversely by means of the other T. F.s. For example tan -1 06 means the least positive angle whose tangent is 06. Note (i) — The origin of this notation seems to have come from considering that, if it is supposed possible that the symbol sin could be separated from its A , and treated as an algebraic entity, we should then be able to write the eq'n sin x = OC oc . . as x = —r- = sin -1 a sin * Sometimes arc sin a. 138 Trigonometry — Chapter VII Note (2) — Since sin OC = sin |n 77 + (— i) n Oi], .'., strictly speaking, sin- 1 (sin (X) = n it + (- i) n a so cos -1 (cos a) = 2 n 77 + (X and tan -1 (tan a) = mr + a* Zk/"' — If an assigned trigonometrical function has a given value, then the least positive angle (expressed by means of this trigono- metrical function) for which this is true, is called the corresponding inverse trigonometrical function. Note — For brevity we shall use I . T. F. to denote the words inverse trigono- metrical function. A s expressed as I. T. F.s are capable of combination in manner very analogous to the combinations of the corresponding T. F.s. The mode of procedure is always uniform ; and, once laid hold of, the manipulation of the I , T. F.s is easy. Thus a table similar to that on p. 28, may be formed for the I. T. F.s. An example will show how any I. T. F. may be ex- pressed in terms of any other I. T. F. : for instance, to express sec -1 06 in terms of cosec -1 06. If sec' 1 06 = 6 sec 6 = 06 a ... sec 6 .'. cosec (7, which = also = Vsec 2 6 — 1 06 Voc 2 — I 06 .*. ft i.e. sec -1 06, = cosec -1 Vol 2 Proceeding in this way we get the table here given : the Student should verify this table. * For a full development of this idea see Chrystats Algebra, vol' II, pp. 235-238 ; where also the graph of sin -1 x is drawn. o o CO o •-*■ O CO CO CD o JB o ~" CD O iH- 3 CO 3 o 1 1 | 1 1 1 h-« I-* M M M ft ft ft ft ft ft CO CO CO C0_ 3 3 2 3* CO 1 1 i l u> . M M CO 3* 3 1 < < < i-i CD CO O 1 < I, sin -1 a and cos -1 (X are imaginary, and „ < i, sec- 1 OC „ cosec" 1 a „ „ ; but tan -1 a and cot -1 OC are always real. Exercise Fill in an 8th column and row with the corresponding values for vers -1 OC So again we may combine two (or more) I. T. F.s into a single I. T. F. of the same kind ; or we may prove a formula for I. T. F.s analogous to any formula for T. F.s ; or we may solve an eq'n ex- pressed in terms of 1 . T. F.s ; or we may by means of the I . T. F. notation write a symbolical expression for the general solution of a trig'l eq'n, when we have found from it a value for a T. F. which does not give any known A ; or, finally, we may by means of I. T. F.s find symbolical expressions for the roots of many alg'l eq'ns. Note — Obviously, from the very meaning of the CO- in cosine, sin -1 6 + cos -1 = 77/2 .*. sin- 1 = 71/2 — cos -1 Sim'ly tan- 1 6 = nh — cot- 1 ^ and sec -1 6 = 77/2 — cosec -1 Q These transformations are sometimes useful. Examples 1 . To express tan -1 OC + tan -1 /3 as a single similar I , T. F. If x = tan -1 OC, and y = tan -1 (3 then tan x = OC, and tan y = /3 tan x + tan y tan (x + y), which = also = 1 — tan x tan y' oc + (3 1 -a/3 •\ x + y, i.e. tan- 1 a + tan -1 /3, =tan _1 ^r J ' 1 — OCp Examples 141 Note — This formula may be taken to mean — any A whose tan is OC + any A whose tan is /3 = some one of the A s whose tan is -= 1 -a/3 2. To express vers -1 a + vers -1 /3 as a single similar I. T. F. If vers -1 a = x, and vers -1 /3 = y then vers x = (X, and vers y = /3 .*. cos x = 1 — OC, and cos y = 1 — /3 .*. sin x = V2 OC — OC 2 , and sin y = V2 /3 — /3 2 .'. cos (x + y), which = cosx cos y — sin x sin y, also = (1 - OC) (1 - (3) - V(2 OC - OC 2 ) (2 {3 - (3 2 ) .-. vers (x + y) = OC + /3 - 0C(3 + V{2 oc - oc 2 ) (2 /3 - /3 2 ) x + y, i.e. vers -1 OC + vers -1 ^, = vers- 1 {0C + L3-OCJ3 + V( 2 oc - oc 2 ) (2 (3 - /3 2 )} 3. Toshowthat the three angles, tan -1 OC, tan -1 8, tan -1 — — — ^-^ 75 I + OC + p — OLp together make up half a right angle. Using the formula of Example 1, we get sum of A s ■ ten -i«±4 +te n-i I - g -g- a g" i - oc(3 1 + oc + (3 -ocf3 a + /3 1 - a - /3 - a/3 = tan- 1 '"^ i+a+/3-a/3 (a + /3) (1 - oc - (3 - a/3) (1 - oc(3) (1 + a + (3 - a/3) -a/3 + a 2 /3 + a/3 2 + a 2 /3 2 _ i a + a 2 + a/3-a 2 /3+/3 + a^+/3 2 -a^3 2 + i-a-^-a3 = tan i+a + /3-a/3-a/3-a 2 £-a/3 2 + a 2 /3 2 -a + a 2 + a/3 + a 2 /3-/3 + a/3+/3 2 + a/3 2 , 1 + a 2 + /3 2 + a 2 /3 2 — tan- 1 — — ~ Ian i + a 2 + /3 2 + a 2 /3 2 = tan -1 1 = half a right A 142 Trigonometry — Chapter VII 4. To find a formula connecting I. T. F.s, analogous to the formula sin 3 OC = 3 sin a — 4sin 3 a If sin a = x then a = sin -1 x and 3 (X = 3sin _1 x But 3 a = sin -1 (3 sin a — 4 sin 3 a) .-. 3sin _1 x = sin -1 (3X — 4X 3 ) which is the formula req'd x x 5. If sec -1 - + sec -1 a = sec -1 r- + sec -1 b, to fond x in terms of a and b. x x Putting OC, /3, 6, (p for sec -1 a, sec -1 b, sec -1 - , sec- 1 7-, respectively a d we get sec oc = a, sec/3 = b, sec0=-, sec = r-; and 6 + OC = (j> + /3 .-. cos cos a — sin 6 sin a = cos r ig 2 ( nVs j 2 where r is any integer, pos', neg', or zero. 7. ^SWztf, £y £#£ aid of trigonometry , the equations Vx (1 - y) + Vy(i - x) = a) Vxy + V(i - x) (1 - y) = b) where a d^af b «r = a^ sin 6 sin cj) + cos cos (j) = b ) -(/)) = b ) •'• Q - \ {sin- 1 a + cos -1 b} = l {sin _1 a — cos -1 b} r. /sin^aX /cos- 1 b\ „ /sin-^X . /cos- 1 b\~i 2 ... x = [_s,n (— ^J cos (-^— j + cos (-!— jsm (— ^)J f . /sin _1 a\ /cos -1 b\ /sin- 1 a\ . /cos -1 b\~i 2 y = |_ sin (__) cos (-5—; - cos (— j—) s.n (-^-)J Note— If we give to a, b the numerical values of known sines and cosines, the results will be verified. For example, it a = — - = b, so that sin -1 a = 6o°, and cos -1 b = 30 , then x = f, and y = 4 which values will be found to satisfy the eq'ns. whence sin (Q + (f>) = a cos (6 — (/)) 144 Trigonometry — Chapter VII Exercises Prove that — 1. tan -1 !- + tan -1 -! = — (Ruler's formula) 4 2. cot- 1 ! + cot- 1 \ = i35° 3. sin- 1 -— + cot- 1 3 = 45 4. tan- 1 1 + tan- 1 i + tan- 1 \ + tan- 1 § = 45 IT 5. 4 tan -1 i — tan -1 -gig = — (Machiris formula) 4 6. tan- 1 ! -cot- 1 ! = cot_1 -X 7. 2tan- 1 T ^3-tan- 1 r ^ = tan- 1 2j9 « 1 9 ■ 1 4 T 8. cos- 1 — j==. + sin- 1 — p^ = - V82 V41 4 9. tan (tan- 1 -! + i\ax\~ 1 \ + tan- 1 ^ " ^) = Tnnrr 10. sin- 1 V — — = tan- 1 V - 6 + OL OC 11. sin- 1 x + sin~ 1 y = sin- 1 (xVi — y 2 ± y V(i — x 2 )) 12. cos -1 x + cos~ 1 y = cos -1 (xy + V(i — x 2 )(i — y 2 )) 13 . tan-! ( """M - tan- ( <* ~ S '" 6 ) = \i-asinO/ V cos0 / - - , a + 8 cos x , a — 8 cos x 14. cos- 1 — £— - + cos- 1 -5 — £ (3 + a cos x /3 — a cos x (S -i a 2 sin 2 x- a 2 + /3 2 15. i tan- 1 [2 tan {a + tan- 1 (tan 3 a)}] = a 16. . tan- 1 i n/^3 tan ?i = cos-, (* + « cos ^ (. a + /3 2) Va + /3cos<9/ 1 ** , 1 — x 2 ,i-y 2 _ 2 (x — y) (1 + xy) 17. cos- 1 s - cos- 1 ^ = tan~ 1 r — ^- ^ 1 + x 2 1 + y 2 (1 + xy) 2 - (x - y) 2 a 2 sin 2 x + a 2 - /3 2 = cos -1 — — '— Exercises 145 „ _ 1 0L-, - Of 2 . Of 2 - Of 3 . « n -l - a n 18. tan- 1 — + tan- 1 2 „ J + &c + tan- 1 - — — 1 + a x a 2 1 + of 2 of 3 1 + a n _ x of n = tan -1 OCi — tan -1 Of n ,« . , a«i/3i . . 2CV 2i 8 2 . 2 Of n /3 n 19. s n- 1 -^— + sin- 1 — - — ^— a + &c + sin- 1 — -= 75 - l oc^ + ps of 2 2 + /3 2 2 + sin- 1 (Of sin$) then (j) = 6 + tan- 1 (^-| tan fl) 27. If cos- 1 - = 2 sin- 1 ^> then Of 2 = Of x + 2 y 2 a a 146 Trigonometry — Chapter VII 28. If x 2 = a 2 + b 2 + ab, and tan 2 (j> = 2 cosec (2 tan -1 — J then x = Vab . sec (f) 29. If cos- 1 - + cos- 1 t- = Ot, a b then 0Y - ^ cos a + (|Y = si n 2 a Sandhurst : '88. 30. If tan- 1 J + 2 tan- 1 i + tan- 1 f- = - - cot- 1 x then x = ||| 31. If tan- 1 - + tan- 1 -^ — = tan- 1 x (x 2 — x + 1 a — 1 then x = a, or a 2 - (X + 1 32. The algebraic equivalent of sin -1 x + sin -1 y + sin -1 z + sin -1 u = n tt, where n is any integer, is {4 (s - x) (s - y) (s - z) (s - u) - (xy + zu) (xz + yu) (xu + yz)} . x J4s(s— x— y)(s— x— z)(s— x — u) — (zu— xy)(yu— xz)(yz— xu)} = o where 2S = X + y + z + u Math' Tri': '88. 33. If a = b cos + c sin 9 and b = a cos 6 then either a 2 — b 2 = o and sin 6 or a 2 + c) 2 = b 2 + c 2 + c sin 6 ] + c) sin ) and Q = tan- 1 ^ + tan- 1 £ a b 34. If tan 3 6 = tan (0 - OQ then 6 = J {(X + sin- 1 (3 sin OC)} where, unless OL ^ sin -1 ^, there are no real solutions. Q. C. B. : '69. Exercises 147 35. If tan (n cot 0) = cot (n tan 0) then = ~ + (- i) m J sin- 1 4 n/(2 r + i)w /^ CVurc^: '60. 36. If cos 2 cos 2 (X + 4 cos sin (0 - a) sin 3 a = sin 2 OC cos 2 a then = n 77 + tan -1 (2 tan a ± cot a) 37. Find the value of x in each of the following — (1) tan^x + Jsec- 1 5X = - 4 IT (2) cot- 1 (x - 1) - cot- 1 (x + 1) = — 12 (3) vers -1 x — vers -1 ocx = vers -1 (1 — OC) (4) vers -1 (1 + x) - vers- 1 (1 — x) = tan -1 (2 Vi — x 2 ) / \ - I — X 2 , 2X 4 77 (5) cos- 1 — — a + tan- 1 I + X^ I — X I 7J- (6) 2 sin -1 — — = -£ + chd- 1 Vx 2 — 2 x + 2 " Vx 2 + 2 x + 2 OC where chd OC (contracted for chord OL) is defined as 2 sin — • 38. The following equations require the inverse trigonometrical functions to express their complete solutions : find the solution in each case : 0, (j) are always the unknown angles ; others are supposed known — (1) V2 sin 2 = cos 3 (2) cosec + cosec 3 = 4 cos 2 - cosec 2 2 sin OL cos (/3 + 0) tan /3 <3) «">0co.«x + fl) " SJJa Jftfl , ^ : , g2 (4) 35 sin 3 + 20 cos 3 + 39 sin — 20 cos = o Pelf and Peml/ , Caml/ : '79. (5) tan (0 + a) tan (0 + (3) = tan 2 (6) (1 + cos OL cos 0) tan (0 + a) + (cos OC + cos 0) sin OC =0 (7) m sec 2 tan (a - 0) = n tan sec 2 (a - 0) (8) sin (77 cos 0) = cos (77 sin 0) L 2 148 Trigonometry — Chapter VII (9) tan (0 - OC) tan (0 - /3) = tan 2 (10) p sin 4 — q sin 4 <|) = p ) p cos 4 — q cos 4

) = T V sin 2 cos 2 $ + cos 2 sin 2

+ cos0 = /3 ) (16) cos 7 + 7 cos = o (17) tan a tan = tan 2 (a + 0) - tan 2 (a - 0) (18) tan 2 = 2 tan a tan (3 sec + tan 2 a + tan 2 /3 (19) tan 2 + sec 2 20 = - — 3 ~ IO Vl Show that one of the solutions is nearly tt/S- (20) 6 cos 3 — 10 cos 2 + 5 sin 20+22 cos — 5 sin = 10 iHfa/A' Tri'\ '84. 39. Show that 2 (X is one value of , 1 — a 2 cos 2 OC — 2 a sin a n cos 2 a + 2 a sin a — a 2 cos -1 s -. — COS" 1 + a 2 — 2 a sin OC 1 + a 2 — 2 a sin OC Ox f First Public Exam': '79. 40. Show that the sum of any number of terms such as a sin (0 + OC), b sin (0 + (3), c sin (0 + y), &c can always be reduced to the form x sin (0 + (j)) where x 2 = S a 2 — 2 2 (ab cos OC — (3) and ) + sin -1 (sin 6 — sin (p) is a right angle. 46. Two circles (radii R, r) cut at an angle OC ; prove that the area com- mon to them is (R 2 - r 2 ) tan- 1 rsm0C + r 2 q _ Rr sin Q R + r cos a Find also the length of their common chord. 47. If Of, /3, y, &c are the roots of the equation x 11 - Pl x 11 " 1 + p 2 x n ~ 2 - p 3 x 11 " 3 + &c = o show that tan- 1 (X + tan- 1 8 + tan~ 1 y + &c = tan- 1 Pl ~ Ps + Ps - o &0 ' 1 - p 2 + p 4 - &c 48. In a triangle ABC, if S is the circumcentre, I the in centre, O the ortho- centre, M the mid point of BC, and L the mid point of the altitude of A ; show that the angle made by BC with — i°, ML is cot- 1 (cot B ^ cot C) 2°, IS „ tan -1 cosB + cosC — i/sin B ^sin C 3% SO „ tan- 1 tan B tan C - 3/tan B ^ tan C _ / B ~ C / B 4°, 10 ,, tan -1 (cot cos A/ 2 sin — sin C . B ~ C > sin 2 2 / Wolstenholme : 533-6. CHAPTER VIII Elimination § 4-3. Def — As in Algebra, when from a given set of equations we deduce a new equation, not containing one (or more) of the letters originally involved, we are said to eliminate such letter (or letters) between the given equations, and the new equation is called the eliminant (or resultant) of the given equations with respect to the letter (or letters) got rid of; so also in Trigonometrjr, when from a given set of equations we deduce a new equation, not con- taining one (or more) of the angles originally involved, we are said to eliminate such angle (or angles) between the given equations, and the new equation is called the eliminant (or resultant) of the given equations with respect to the angle (or angles) got rid of. To acquire facility in eliminating between eq'ns (whether in Alg' or Trig') is of the first consequence in a mathematical training ; for, in almost all mathematical problems, treated analytically, the solution consists mainly of two processes — r i°, writing down eq'ns ; and, 2°, eliminating between them. Now this facility comes from observation of methods, and practice in applying them. No rules of general application can be given ; but certain methods and artifices are frequently useful ; and should certainly be familiar to the Student. Of these some Examples, which should be carefully studied, will now be given. In effecting a trig'l elimination we are, of course, at liberty to make use of any identity that can be usefully pressed into our service. Thus, if we can reduce two eq'ns to the forms sin 6 = a cos 6 = b 152 Trigonometry — Chapter VIM then, squaring and adding, and recollecting that sin 2 + cos 2 = i we get a 2 + b 2 = i and is eliminated. This artifice is of constant use. So also, if eq'ns are reduced to the forms sec = a \ tan = b ) then, squaring and subtracting, and recollecting that sec 2 - tan 2 = i we get a 2 — b 2 = i as the eliminant of the eq'ns with respect to 0. Or suppose that we are given the eq'ns i — cos = a sin 6 = a sin u i = bsin<9 ) i + cos 6 and it is req'd to elim' 0. Multiplying the corresponding sides of the given eq'ns together, and recollecting that i — cos 2 = sin 2 we get ab = i as the req'd eliminant. Again the relations between the roots and coeff's of an eq'n can often be advantageously used. E.g. if a sin + b cos = c and a sin + b cos

V — cost? + f-sin 6 = i * a b coscj) + j- sin$ = i cos0coscf) sin sin between the equations x cos 6 + y sin 6 = 2 a \ x cos (p + y sin (£ = 2 a (f> 2 cos — cos — = I 2 2 From the two 1st, we have, by well-known algebraic processes, x y 2 a 6 + 2 cos — COS — 2 2 = x + 2 a, by 3rd eq'n Examples i55 .\ x 2 + y 2 — (x + 2 a) 2 .«. y 2 = 4 a (x + a) is the eliminant. /a 2 + sin a sin

/a 2 + sin /3 sin (J)/b 2 + i = o ) where a 2 + b 2 = i, produces cos a cos/3/a 2 + sin a sin /3/b 2 + i = o iJSfa/A' Tri\ '79. The following solution was given by Prof Cayley in the Messenger of Mathematics, vol' v, p. 24 (old series) cos $/a 2 : sin

sin^> ax by = c 2 \ = ^2 coscp' sin<^/ > where c 2 = a 2 — b 2 cos COS 4> +

' = c 2 (sin (j> — sin <£>') cos — cos cf)') sin ' by cos- — — = - c 2 sin -*- — -*- «cos 2 - — — - cos 2 - — - t J 2 2 ( 2 2 ) ,. by ^o=-c 2 ^40(1.30 ... by = — v'c 2 O - a 2 Q 2 ' c .-. c 2 y 2 + a 2 b 2 Q 2 = b 2 c 2 Q ....... (B) .'. , substituting for 12 from eq'n (A) in eq'n (B) we have as the eliminant (a 2 + c 2 ) 2 y 2 + a 2 b 2 (x + c) 2 = b 2 c (a 2 + c 2 ) (x + c) Note — This gives the Locus of the cross of normals at the ends of a focal chord of the ellipse (x/a) 2 + (y/b) 2 = 1, (j), (j)' being the eccentric /\ 8 of the ends of the chord. The same result is found, by a quite different method, as a solution of the Problem, in Salmon's Conies, 6th ed f , p. 211. Exercises 1. Eliminate between the equations — (1) tan + sin = a J tan 9 — sin = b ) (2) sec + tan = a } cosec + cot = b ) (3) cosec — sin = a } sec 9 — cos0 = b ) (4) sin 9 + sin 2 9 = a } cos 9 + cos 2 = b } (5) cosec tan 3 9 (cosec 2 9 + 1) + sec = a J tan 3 9 (cosec 2 + 1) - tan = b J (6) sin (0- a)/ sin (0-/3) = a/b £ cos (0 - a)/cos (0 - ft = a'/b' ) Exercises 159 (7) sin cos (cos — sin 0) = a ) sin cos (cos + sin 0) = b ) (8) 3 cos (0 + a) - 2 sin (0 + a) = cos (0 - a) 3 cos (0 + /3) + 4 sin (0 + (3) = cos (0 - /3) -a) J -J3) ) (9) tan a = (1 + m) tan 0/i - m tan 2 ) tan /3 = (1 - m) tan 0/i + m tan 2 * (10) sin cot x = sin (0 + OC) cot y = sin (0 + a + /3) cot 2 (11) (a- b)sin(0 + a) = (a + b) sin (0 - a) a tan- = btan- + c 2 2 (12) a sin + b cos = c = a cosec + b sec (13) x = a (1 + sin 2 cos 2 0) } y = a sin 2 sin 2 ) (14) a sin 2 + b cos 2 = c } a cosec 2 + b sec 2 = c) ) (15) sin (0 + a) = k sin 2 J sin (0 + /3) = k sin 2 ) (16) cos (0-ol-P) cos (0-/3) = m = cos (0- a + /3) cos(0 + (3) (17) a cos + bsin0 = c = a cos 3 + b sin 3 Ox' 2nd Public Exam' \ 'go. (18) x sin + y cos = a Vi + sin 2 cos co x cos - y sin = a cos 2 cos o)/^i + sin 2 cos co Ox' 1st Public Exam': '77. 2. Eliminate and (f> between the equations — (1) x=acos m 0cos m (£ ) y=bcos m 0sin m <£ z = csin m i6o Trigonometry — Chapter VIM (2) cos 2 = cos a/cos j3 * cos 2 (j) = cos y/cos (3 - tan 6 /tan <£ = tan a/tan y , (3) a cos 2 + b sin 2 = m cos 2 <|> a sin 2 + b cos 2 = n sin 2 <£ m tan 2 = n tan 2 <£ (4) a cos + b sin = c a coscf) + b sin (j) = c tan tan <£ = m , (5) a sin

) = c sin cos — cos <£ = 2 m (6) sec — sec

) = o (9) sin + cos (j) = 4 mn cos + sin (j) = 2 (m 2 — n 2 ) + . 6 + , , „ cos ! — sin L = y 2 (nv — n 2 ) Ox' Jun' Math' SchoV: '78. Exercises 161 3. Eliminate (f) from the equations x m y ■ jl - cos (p + ^sm cp = i cos (p sin (p Note — These are the eq'ns to the tangent and normal to an ellipse at the p't whose eccentric /\ is (f) : the elimination gives the Locus of their cross, which is of course the ellipse itself. 4. Eliminate <£ from the equations ax/cos (£> — by/sin + by cos = — - — - tan t i — e 2 and tan [ — + — ) = — tan (— + - ) V 4 2 / i - e \4 2/ show that sin = - e 7. Eliminate between the equations a cos 2 + b sin 2 = c a' cos 2,0 + b' sin 3 + b sin 20 = c } + b' sin 30 = o ) Note — Put xfor tan 0, and elim' between b'x 3 + 3a'x 2 -3b'x-a' = o \ and (a + c) x 2 — 2 b x + c — a = o ) M John's CamV\ '87. \6% Trigonometry — Chapter VI 1 1 8. Show that if is eliminated between the equations 6 tan (0 + OC) = 3 tan (0 + j3) = 2 tan (0 + y) the eliminant can be arranged as 3 sin 2 (a - /3) + 5 sin2 C/3 - 7) - 2 sin 2 (y - a) = o /V/' C anil/ 1 '72. 9. Eliminate from the equations x (1 + sin 2 - cos 0) - y sin (1 + cos 0) = a (1 + cos 0) | y(i + cos 2 0) -xsin0cos0 = asin0 ) Note — Solve for x and y. Wolstenholme : 416. 10. If a cos 3 + b cos 2 + c cos = o | and a cos 2 + b cos + c = o J eliminate ; and show that = m 7T, if a and C are unequal. 11. Show that if x, y, z are eliminated between the equations x = y cos y + z cos /3 y = z cos OC + x cos y z = x cos /3 + y cos a the eliminant can be arranged as II sin 2 OC = IT (cos OC + cos (3 cos y) Clare Cam//\ '74. Note — Use Sarrus' rule: then mult' by cos OC cos /3 cos y, a# = b/x cos (0 — (f)) = c/x 13. Eliminate and <£ between - cos + -r cos = 1 = - sin d + r- sin a b ^ a b r and ( - J — ( jM = \j? (sin 2 — sin 2 (£) ii. 7 1 . xxxiii. Exercises 163 14. Show that the elimination of /3 between x 2 cos OC cos (3 + x (sin a + sin /3) + 1 = o and x 2 cos /3 cos y + x (sin /3 + sin y) + 1 = o produces x 2 cos y cos a + x (sin y + sin a) + 1=0 E. T. xlv. ilfarfA' 7V/: '69. John's Camb'\ '81. 15. Show that an eliminant of 6, between Q cos0 + f: sin = 1 \ a b Christ's Camb'\ '49. x y - cos (b + j- sin 6 = 1 a ^ b ^ - d> OC-6 OC-d) 4 cos COS COS = I 2 2 2 produces ( cos OC) + ( ~ — sin OC) = 3 17. Eliminate /: '40. Jfa/A' Tri'\ '71. 164 Trigonometry— Chapter VI 1 1 19. If tan y/tan /3 = sin (x — (X)/sin (X \ and tan y/tan 2 /3 = sin (x — 2 a)/sin 2(X) show that tan x/sin 2 Oi = i/(cos 2 a — cos 2 /3) yJ/a^' 7W: '74. 20. Eliminate between a cos x cos (j) + b sin x sin <£ = c 1 a cos y cos (£ + b sin y sin (f) = c 3 21. If x = A.acos<£, y = A. b sin <£ and - cos

)/(a 2 + b 2 ) a " (^0=g) 2+ gd 2 Note — 77zw is a solution of the Problem to find the Locus of the centres of all rectangular hyperbolas having contact of the 3rd order with the ellipse (x/a) 2 + (y/b) 2 = 1. 22. Eliminate 6 between x cos 6 + y sin 6 = 2 a sin 20 ^ y cos 6 — x sin = a cos 2 ) Note — The elimination gives the Envelope of normals (i.e. the evoluie) of 2. 2. 2. the hypocycloid x 3 +y 3 = a 3 . 23. Eliminate (p from x v - cos (/) — ^ sin (£ = cos 2 cf) - sin 9 + ~ cos 9 = 2 sin 2 9 ) Note — The elimination gives the Envelope of the chords of curvature of the ellipse (x/a) 2 + (y/b) 2 = 1. 24. Eliminate 6 and (f) between a tan 3 + b tan 6 = c a tan 3 <£ + b tan (j) — c 6 + (b = it/ 4 7 i?. 27. •SV^/': '90. Exercises 165 25. Eliminate between a cos 3 + b cos 2 + c cos + d = o a cos 3 <|) + b cos 2 + c cos (|) + d = o cos + cos (f> = m 26. Eliminate between x cos (0 + a) + y sin (0 + a) = a sin 2 1 y cos (0 + OC) — x sin (0 + a) =■■ 2 a cos 2 0} -ff. £/". ^^/: '87. 27. Show that if x, y, a, b are eliminated between the equations ©'♦©■- x = r cos 0, y = r sin a 2 + b 2 = r 2 + K 2 , ab = rK sin OC the eliminant may be arranged either as + cot 2 = cot 2 a + / -j \ cosec 2 a or (-\ sin 2 = sin 2 (OC ± 0) 28. Eliminate from the eq'ns p sin (a — 3 0) = 2 a sin 3 jo cos (a — 3 0) = 2 a cos 3 r. C. Z>. : '75. 29. Show that, unless sin 2 (X = o, the result of eliminating and (p from the simultaneous equations x cosec + y sec = 1 = x cosec (j) + y sec y and - ) + b cos (0 — +)() + b cos ((j) — \) + c = o a cos (^ + 0) + b cos (^ — 0) + c = o , where 0, + (a- b) 2 sin 2 <|>} x 2 + 2 (a + b) ccos^.x + c 2 - (a- b) 2 sin 2 c|> = o .•. cos COSX {and sim'ly sin sin ^) are known: substitute in 3rd eq'n. cos 2 sin 2 fl _ 1 33. if —^r + -^2- - p2 r sin = p s\n

■ b 2 ~ a 34. Eliminate between a sec + b cosec = c 1 a sec — b cosec = cos 2 0) Joh' Camb' : '42. 35. Eliminate x sin 3 $ + y cos 3 (£> = b cos (<|> + j- \ Christ 's Emm' and Sid' , Camb': '81. Exercises 167 36. Prove that the eliminant of between 4 (cos OC cos + cos $) (cos OC sin + sin (p) = 4 (cos OC cos + cos y) (cos OC sin + sin y) = (cos (J) — cos y) (sin $ — sin y) can be written jcos ((f) — x) — cos 2 a} -[cos ($ — yj) + i} = o Math' Tri'\ '73. Caius CamV: '74. 37. Show that the elimination of x, y, 2 between the equations a/sin x + y + b/cos x + y = c/cos x — y a/sin y + z + b/cos y + z = c/cos y — z a/sin z + x + b/cos z + x = c/cos z — x where no two of x, y, z are equal, or differ by a multiple of 7T, produces a 2 + b 2 = c 2 R. U. Schol'-. '88. 38. Eliminate between sin 3 + m sin a + + 2 sin @ — sin y — sin /3 + y + = +y +0=0 ] + y + = o ) cos 3 + m cos a + + 2 cos (3 — cos y — cos /3 + y iH/kM' 7W: '87. Note — Mult f 1st eq f n by sin 0, 2w ch'd ANB, but < AT + BT .'■. arc AX > NA, but < AT AX NA , AT > ^-r, but < CA" CA CA Some Important Limits 169 .-. > sin ft but < tan i.e. sin 0, 0, tan are in ascending order of magnitude , sin# tan •'• &1SO — , I, -T— ,, ,, ,, „ s\nO s\nO n or —7j-> ii —q- •sect' Now, as diminishes, sec continually approaches (and ulti- mately has) unity for its Limit. approach equality, as is diminished : e ' ' e in6 /tan 0" and 1 ■ / /sin0\ 1 • / / tan ^ Cor'(i)— lfx° = r 'sin 0\ 7T JT_ 180 So also, if y" = r (|) .. , /sip x\ ,. , / sin fl x ,. , /sin 0\ 7T 1 Lim' (**H2\ - Lim' / sin ^ \ L,m y = o^ y / |- Lim ^ = / l8o x 60 x 60 1 77 . , / sin (j) \ 7T 7T - Lim ^ = 0^— ^— J 648oo - 648oo CVr. (2) — By the formula at end of p. 64, we have a a a „ 7r sin a COS — COS — COS — &C COS — ; 2 4 " 8 2 n n . a ^ z 2 n sm — 2 n Now 8,n * -(™*\ ** . a in — 2" „n n ^ \ sin 2 170 Trigonometry— Chapter IX And, by the Theorem, if OC is in radian measure, and n is increased in- definitely, Lim' ' n = 00 . OC sin — 2 n = 1 '. Lim sin OC % sin OC \ 2n sin-J , . . / a a a *. Lim' ^ ( cos - cos - cos — &c cos H=«5 ^248 Put 7r/2 for OC, in this formula, and we get V 2 V 2 + V 2 'V 2 + V 2 + V2 K\ _ sin a {Ruler's Formula 2 TT 2 Sec ad infin' (Vieta's Formula) This gives another method for approximating to 7T. § 4-5. Theorem — If 6 radians is a positive angle less than TV h, then tan 6 — 6 > 6 — sin 6. Let C be the centre of a O ; 2 6 the inclination of two of its radii CP, CQ ; PL, QL the tang's at P, Q ; and A, B the respec- tive mid' p'ts of PL, QL. Let CL cut PQ in N, AB in R, and the O in X. Then RA is the radical axis of a system of which the O is one, and L is a limiting p't *. * See Euclid Revised, pp. 349, 350. Some Important Limits 171 .-. AB will not meet the O .•'. PA + AB + BQ > arc PXQ PL NP / arc PX \ ••• CP + CP > 2 ^ CP ; i.e. tan + sin > 2 i. e. tan — > — sin ft if is pos' and < irfe * §46- Though, when is a small positive angle, sin ft ft tan are nearly equal; yet, since then tan — 6 > s\n — ft the approximations tan = 0, COS = i, are not so close as sin = 0. /j sin 2 # .But, since 1 — COS u = 1 + COS0 in 2 \ 0* ,. Lim^. (.-oosfl)=Lim'^ (-^_)=: ; .-. cos = 1 ■ j nearly 2 / 2 \~ From this tan = — = ( 1 ) + 1 2 3 l i. e. tan = -\ 3 nearly 2 # 2 3 .'.ft 1 , H are respectively approximations to sin ft cos ft tan ft when is a small positive angle. We proceed to get still closer approximations. * This proof is due to Prof 7 Genese : E. T. xlii. JJ2 Trigonometry — Chapter IX § 4-7. Theorem — If radians is a positive angle less than 7r/2, then si n lies between and — 3 l\ ', COS 6 lies between 1 — 6/2 and 1 — 2 h + 0*/i6; and tan lies between + I2 and 6 + 3 / 4 . For sin 0=2 tan - cos 2 - 2 2 and .*. > cos 2 - , when < w/2 and pos' i.e. > ( 1 — sin 2 - j .*. , a fortiori, >^] I ~( _ )f i. e. > - s / 4 And it has been shown (§ 44) that sin < 0, .'. sin lies between and — h, when is pos' and < irfa. Again cos# =1 — 2 sin r »2 _ 2 '0 6 2 /2 > I - 2 /2 Also cos U < 1 — 2 < V / 1 ... < ! _ 2 / 2 + 0*/i6 .*. COS0 lies between i — 2 fa and 1 — # 2 /2 + #Vi6, when is pos' and < irh. Q . sin0<0 ) s.n^>0- 7 Smce, 10, ^ [ and} 20? ^ ^ cos > 1 cos # < 1 V —7 2 } 2 10 Some Important Limits 173 ', *-? .-. , io, tan 6 < -p ; and, 2°, tan 6 > j 2 — -^ 1 1 + -7 2 2 16 / 2 \ _1 i.e. tan#< 6 I 1 - y ) buttan0> ^- T )(i- T + -) 6 s 6 3 i. e. tan 6 < 6 + — ; but tan > 6 + — 2 4 s So that tan # = # H is a good approximation. 3 Hence, for small values of 6, we have that sin<9 = 0-- 4 COS = I 2 tan 6 = 6 + — 3 are good approximations, if powers of higher than 6 are of no consequence. We can however go closer still to sin 6 without taking account of powers of 6 higher than 6 . § 4-8. Theorem — If 6 radians is a positive angle less than TV I 2 then sin 6 is greater than 6 — 6/6', COS 6 is less than 1 — 0*fa + 0V24 ; and tan 6 is greater than 6+6/3*. * Rawson : Messenger of Mathematics : III. 101. 174 Trigonometry — Chapter IX Since sin = 3 sin 4 sin 3 - 3 3 .-. 3 sin - = 3 2 sin -5 - 3 x 4 sin 3 -j „ . 6 o.O o . , 3 2 sm -3 = 3 3 sin — - f x 4 sin 3 — 3 3 3 &c &c n-1 • & n ■ Q n—l • * 3 sin -^^ = 3 sin — -3 X4Sin 3 — 3 3 3 .*. , adding corresponding sides, and omitting terms common, . /\ XL ■ V ( • 1 ^ ■ * ^ c, . » s\n0 = 3 sin — - 4 jsm 3 - + 3S m 8 - s - + 3 2 sm 3 - F + q ( 3 3 3 &C + 3 n - 1 sin 3 |} But sin ( r 0) < r f/0\ 3 / 6 n-l / # ' sin — - .*. sine/ > <9 3 v 3 n ^ .*. , passing to the Limit when n = 00 , we get 4 0' sin (9 > - -^2- i.e, > - 0*/6 Some Important Limits '75 „ .606" Hence sin - > 2 2 48 /. cos^ (i.e. i-2sin 2 ^ < 1-2^-^y 6 2 6" 1. e. < 1 h — 2 24 \ , on account of both the foregoing inequalities, tan 6 > i.e. > •-! 6 2 0' 1 + — 2 24 6 2 6' (*-f)Hf-e 3 Hence we have the following very good approximations — When r 6 is so small an angle that 6' may be neglected, sin<9 = 6- 6 l cos 6 = 1 2 tan (9 = + 6' where, in each case, the dexter expression is slightly less than the true value of the cor- responding T. F. These are the approximations which the Student should recollect for use. Note — Since i° = -oi 745 of a radian ; and that (.01 745)4/24 has no significant figure in the 1st six places of decimals ; we see that the above approximations are true if 6 ^> i°, at least as far as the sixth decimal place. 176 Trigonometry — Chapter IX § 4-9. Theorem — If radians is a positive angle less than 7r/2, then qsin# A , tan#+2sin# —2 -, 0, and - 2 -f cos 3 are in ascending order of magnitude. For jin e . e e 2 sin - cos- 2 2 . 2 sin - 2 2 + COS0 J ' , 1 + 2 cos - sec — h 2 cos - 22 2 But sec — H cos - > 2, since # is pos' and < irh £ 2 si n0 2 + cos < 2 f . sin - 2 2 + cos 2 J \ also < 2' . \ sin - 4 r .-. , by n repetitions of this process, < 2 n 2 + cos - 4) sin — 2 (9 2 + cos — - n; i.e. < f . ^ sin — n 2 (9 V V 2 n y / I 2 V + COS 2 n 6 < - , by increasing n to 00 3 Some Important Limits 177 Again tan 6—2 tan - > 2 (2 sin sin 0) if 2 tan - - 2 1 —tan 2 _ — I . ( v\ > 4 sin - ( 1 — cos - J i. e. if ■ J sin - cos e 1 r^r>4sm . ..0 cos" sin' • 2 @ 2 @ 4 sin -cos 2 - or if — ^ > 4 sin 2 - 1/ 4 e e COS C7 COS or if cos 2 - > cos cos - 4 2 But, as < 7r/2. .-. cos - > both cos and cos - 5 / 4. 2 .-. the above inequality is true .-. tan + 2 sin > 2 ( tan - + 2 sin - 1 .-. also > 2 2 (tan - + 2 sin - ) n / . 0\ -., by n repetitions of this process, > 2 I tan — + 2 sin — J r 0\ ( . #Y i.e. ><9 (9 V 7» ; + 2 sin — 2 _0 y .'. > 3 0, by increasing n to 00 . N ljti Trigonometry— Chapter IX . q 3sin0 and < 2 + cos 6 tan 6 + 2 sin 6 3 when 6 is pos' and < 7r/2 Note — By the above investigations, we see that the smaller 6 is taken the more nearly will these limits approximate to each other. Hence, if we can find the numerical value of sin 6, cos 6, tan 6 (where 6 is small), we shall get Limits between which i: must lie. Thus from Chapter V we have that 7T IT IT sin — = *2j;q, cos — = -o66, tan — = -268 12 12 12 IT -268 + .ci8 -786 /. — < ^— < — < .262 12 3 3 and — >^L>. 2 6i 12 2-906 .'. 7T lies between 3-13 and 3.14 Of course by taking 6 smaller we shall get closer Limits. Examples 1. To find approximate values for sin 1' and cos i'. By an easy numerical calculation it will be found that, if r x = i', then x 3 /6 and x 4 /24 have no significant figure in the 1st ten decimal places. .'. , to that order of approximation, r 3-1416 sin 1/ = x = n , = -00029088 180 x 60 And oosi'-i-l(-j£fi|-Y \i8o x 60/ = I — \ O0002909) 2 = -99999995 2. To find sin 10" approximately. // ^. 10x3-1416 io // < ' , of a radian 180 xooxoo i.e. < -000048481 }) 5> >» Examples 179 .-. sin 10" < .000048481 but > -000048481 — -g- (.00005) 3 .'. sin 10" = .000048481 correct to 9 dec'l places. Cor' — When n ^> 10, sin x\" = radian measure of n" = n x radian measure of \" — n sin \" This result will be needed hereafter. 3. To find the Limit, when = o, of m sin — sin m (cos — cos m 0) s 2 Putting — — for sin 0, and 1 for cos 0, the given fraction becomes m a m0 3 . m 3 3 6 v ' 6 whlch = m m 2/)3 2 2 m 3 — m m = — - f — 5 r = — , the req'd Limit 3 (m 2 - 1) 3 ^ 4. To find the Limit, when n = ao , of ( cos — ) • Put x for it, and take logarithms ; then log x = n 2 log cos- = n 2 log f 1 — — - Y since - is very small ultimately = - n 2 ( — 5 + -— + &c ) \2n 2 8n 4 / = — 2 /2, ultimately -0 2 / 2 .*. x = e N 2 i8o Trigonometry — Chapter IX 5. To find the mean centre of a circular arc. Let AB be an arc of a O, whose centre is C, and rad' 3 r ; M the mid p't, and G the mean centre of this arc. Obviously CGM is a st' line. Also, if M 1} G x are respectively the mid p't and mean centre of arc AM, and, ifM 2 ,G 2 „ ,, „ „ „ „ „ „ AM X , and sim'r notation is continued, then Gj G is _L to CM, G 2 G 1 is ± to CM 2 , &c .•: CG = CG, cos -, if ACB = 2 Q 2 = CG 2 cos - cos - 2 4 6 e 6 and, after n substitutions = CG„ cos — cos - cos — . . . cos — ' n 2 4 8 2 n .-., using Ruler's formula, and noting that Lim / n = 00 CG n is r, ** /sin 0\* we get that CG = r ( — zp ) Exercises 1. If 6° = T 0, find which is the first place of decimals affected by using 1—6 2 /2 for cos 6°, instead of 1 — 6 2/2 + 4 / 2 4 » an( * vef ify your result by a table-book of natural cosines. 2. Find the Limit, when T 6 = 11/2, of each of the following — (1) sec — tan 6 (2) cosec 6 - cot 6 * The above proof is due to Prof Crofton : E. T. xiii. Exercises 181 3. Find the Limit, when x = o, of each of the following — (i) (sin x")/x" (2) (tan x' + tan x")/x" (3) (sinax°)/(sin/3x°) (4) (vers OC x°)/(vers (3 x°) 4. Evaluate Lim' _ (tan 2 — 2 tan 0) /0 s Clare and Caius Camb 1 : 74. 5. When r is small, show that 3 = 2 (sin 2 — sin 0) + tan 2 — tan 0, nearly 6. By putting 77/24 for in the results of § 49, find Limits for 7T. 7. Apply Euler 's formula to show that the Limit of (1 — tan 2 — J ( 1 — tan 2 — j ( 1 — tan 2 ^-J . . . adinfin' is $/tan 9 8. Knowing tan 45 find tan 46 , without tables, and correct to four decimal places. R. U. SchoV\ '82. 9. If sin = 2 165/2 1 66, show that is very nearly /19 th of a radian. 10. If sin 010 — 1 9493/1 9494, show that has very nearly the mag-» nitude of i°. 11. If cos 010 + 0/cos has its least possible positive value, show that 0>V3-i Math' Tri'\ '64. 12. Evaluate Lim' x _ cosec 2 /3x°. log e cosax°. a 13. Defining end as 2 sin — , show that, for a small value of 9 3 = 8chd--chd0 2 14. If cos (0 + x) cos h cos h' + sin h sin h' = cos 0, where x 2 , h 3 , h' 3 may be neglected ; and if h + h' = p, h — W = q, show that the number of seconds in x is, approximately, isini" jp 2 *an -- q 2 cot -J Enclopcedia Metropolitana : Trigonometry. 1 82 Trigonometry — Chapter IX 15. If is measured by the radian, find the Limit, when is indefinitely diminished, of each of the following — (1) (1 - cos 0)/0 2 (2) (0 - sin 0)/0 3 (3) (tan 0- sin0)/sin 3 (4) sin n cos (n — 1) 0/sin (5) {0 + sin - sin 2 0)/( 2 + tan - tan 3 0) (6) (0 + tan -- tan 2 0)/( 2 + tan - tan 3 0) (7) (2 sin — sin 2 0) 2 /(sec — cos 2 0) 3 (8) (0 + sin 2 — 6 sin —J / (4 + cos 0—5 cos —J (9) (tan sin - sin tan 0)/0 7 /0 2 (10) (tan 0/0) ' Note — Some of these are given in Williamson 's ; and some in Greenhilfs Differential Calculus, as Exercises to be done by the Calculus : they will all come easily, by using the approximations at the end of % 48. 16. AB, BC, CD are equal consecutive arcs of a circle: TA, TD are tangents : AB, DC meet in Y : XB, XC are tangents, and are produced to meet AD produced in E, F : if B, C, D move up to A, show that, ultimately, A YAD : A TAD : A XEF = 18 : 27 : 25 W. H. H. Hudson : E. T. xxxiv. § 50. If a trig'l eq'n involves a small A , we may approximate to the value of that A , by means of some of the preceding formulae. The method of proceeding is this — Substitute for the T. F.s of the small A the values given in § 46, or in § 48, according to the smallness of the A , and the degree of approximation required. In the eq r n thus obtained neglect all powers of the A except the 1st; and hence find a rough 1st approx' to the value of the A • Substitute this value in the terms involving the 2nd power of the A ; this will give a 2nd approx' to the A • This 2nd approx' may then be substituted in the 3rd powers of the A , giving a 3rd approx' ; &c. Example and Exercises 183 Example If cot OC sin (C + 6) = cot OC sin C cos (j) - cos C sin , and 6 and are so small that their cubes ?nay be neglected, show that, approximately ', 02

cos C .-.(/)=- 0cota + K 1 -^) - (i-^)[cotatanC = - 6 cot OC + J (0 2 - 2 ) cot a tan C .*., as a 1st approximation, we have 02 d> = - 6 cot a + — cot a tan C ^ 2 Substituting this for (j) 2 , ar.a omitting terms in 3 , 4 , we get cj> = - 6 cot OC + i (0 2 - 2 cot 2 a) cot a tan C 02 = -0 cot OC + — cotatanC (1 - cot 2 a) Exercises 1 . If 6 is so small that its cube may be neglected, and if cos (OC — 6) — cos OC cos (3, show that = _ 2 cot a sin 2 — ( 1 — cot 2 a sin 2 —J, approximately. Joh f Camb'\ '35. 2. If 6 is so small that its cube may be neglected, and if sin (pc — 0) = sin OC cos /3, show that the number of seconds in d is, approximately, /tana . J tan 3 a . d (3\ 2 ( -= — 7 - sin 2 1- : — jj- sin 4 — ) \sin 1" 2 sin r 7 2/ 184 Trigonometry — Chapter IX 3. If 6 and (j) are so small that their cubes may be neglected, and if cos (OC + 6) = sin (j) sin OC cos (3 + cos cf) cos OC show that the number of seconds in is, approximately, 2 sin 2 /3 sin 1" sin 2" cot a Hy triers' 1 Astronomy. 4. If OC and 5 are small, and cos (6 + 8) = sin 2 OC + cos 2 OC cos 0, find 1st, 2nd, and 3rd approximations for b. 5. If cos 6 = T 6, approximate to the value of 6. Note — Since 17/4 = -7854, and cos 45 = -7071, .\ = 71/4 — (p } where Since i' = r (7r/io8oo) = r (-ooo29) very nearly h 2 .*., if h > i r , then — > -000000042 2 h 3 But then sin h ~ h < — o and cos h ~ ( 1 I < — \ 2/24 See § 48 .-. sin h = h ) . , ,. . to more than 7 dec 1 places and cos h = 1 - — j when h > *' Now sin (6 + h) = sin 6 cos h + cos 6 sin h h 2 = 1 f 1 J sin 6 + h cos 6, to 7 places Proportional Differences 187 h 2 .-. sin (0 + h) — sin 6 = h cos # sin 0, to 7 places h 2 Unless # is nearly 77-/2, the term — sin 6 will not affect the 7th dec'l place. .-., in general, A sin 6 = h cos 6, and .\ oc h which proves the principle for sin 6, unless 6 is near 73-/2. But if 6 = 77-/2 nearly, then— i°, h cos 6 is so small that it will not affect 7 figure tables at all ; and, for such tables, the differences are said to be insensible : also, h 2 2 , the term — sin 6 would need to be taken into account; 2 but the diff's will not occur at regular intervals ; and they are said to be irregular ■; and the principle does not hold. Note — The practical application of the principle can be seen thus — The T. F. of an A has the same value no matter in what unit the A is measured. Also the ratio of 2 A s > measured in a^' unit = their ratio measured in any other unit. .'., if A is an A (not near 90 ) measured in degrees and minutes, sin (A + n") — sin A : sin (A + i') — sin A = n" : 60" i. e. diff' of sines for an increment of n" = y- x dirF for an increment of 1/ 00 But difF for 1', corresponding to A, is registered in the tables. Hence the diff for n" can be reckoned. A precisely similar investigation (which the Student should go through) will show that A cos# = — h sin 0, and .-. oc h unless 6 is very small, when the differences become insensible and irregular. 1 88 Trigonometry — Chapter X Hence, with that exception, the principle of proportional parts holds for cosines. Note (i) — The diff's for sines and cosines maybe deduced, each from the other, by taking the complementary A • Note (2) — A consideration of the graphs of the sine and cosine (see § 11) will remind us when their diff's become insensible. § 53, Again A tan = sin h/cos cos (0 + h) = tan h/cos 2 (i — tan h . tan 0) But, since sin h = h \ f to more than 7 dec'l places COS h = 1 — — j wh en h > i' .*. then tan h = h, to at least 7 places .-. A tan = h sec 2 0/(i - h tan 0) = hsec 2 0+ h 2 sec 2 <9tan<9 If < 7T 1 4 the 2nd term may be neglected. And then A tan 6 = h sec 2 0, and .-. oc h So that the principle is true for tangents. But if 6 > 77-/4 the 2nd term may not be neglected. And then the diff's are irregular. The diff's are never insensible. In a precisely similar manner it can be shown that A cot 6 = — h cosec 2 6, and .-. oc h unless < 7r/4, when the diff's become irregular. Hence, with that exception, the principle of proportional parts holds for cotangents. Note — The diff's for tangents and cotangents may be deduced, each from the other, by taking the complementary A • Proportional Differences 189 § 54-. Next, since sec(# + h) = /cos cosh — sin sin h =Jcos0 (1 - ^-htan 0\ = sec (1 + h tan + ^ + h 2 tan 2 6\ h 2 .*. A sec = h sin sec* 6 h sec (1 + 2 tan 2 0) When is neither near o or it h the 2nd term may be neglected, and then A sec 6 = h sin 6 sec 2 0, and .-. oc h so that the principle holds for secants. But when is nearly o the dififs are irregular and insensible ; and when 6 is nearly ir 1 2 they are irregular though sensible. Sim'ly it may be shown that A cosec 6 = — h cos cosec 2 0, and /. oc h ; unless is nearly o, when dirFs are irregular though sensible • or unless is nearly irh when they are irregular and insensible, § 55. Hence if/(#) is any T. F., and 8 an increment of 0, then f(0 + 5) -/(0) oc 5 excepting in particular cases : cf. § 60. Note (1) — The particular cases will be best recollected by thinking of the graph of each function. Note (2) — The two terms got in each case for A /(d) can be at once re- covered, by anyone who knows Taylor's Theorem in the Differential Calculus, from the consideration that /(0 + h) =/(0) + h/'(0) + ^/"(0) + &c 19° Trigonometry— Chapter X Examples 1. Given that sin i8° i'= '3092936, find sin i8°o' 23^, without the aid of a table-book. 00 V5 — 1 1-2360679 sin 18 = —2 = — 5 L* = .3090169 4 4 .-. diff r for 6o // = .0002767 •'• „ „ 23" = fo x -0002767 = -0001060 .-. sin i8°o' 23" = .3091229 2. Given that cos (X = 'tfuygg, find OC by the aid of a table-book. From the tables cos 68° 13' = -3710977 and diff for 60" = .0002701 Also if OC = 68° 13'- 8" difF for 8" = -0001022 ... 5" = Z^i x 60- 2701 = 22"-7 .-. a = 68° 1 2' 37"-3 Afr&? — Notice that for A s between o° and 90 , the diff 's are to be added for the sine, tangent, and secant; but subtracted for the cosine, cotangent, and cosecant. This is evident from the consideration that, as the generator of an A passes thro' the 1st quadrant, the T. F.s beginning with co decrease, while the others increase. Exercises 1. Find sin 37 23' 47", by the aid of a table-book. 2. If cos OC = -8241657, find OC by the aid of a table-book. 3. Given that sin 30 1' = .5002519, find sin 30 1/ 17" without using tables. 4. Given that cos OC = -4996532, and that -0002519 is the difference for 1/, find OC without the aid of tables. Proportional Differences 191 To prove the principle of proportional differences for each of the logarithmic T. F.s. h 2 § 56. sin (0 + h)/sin = 1 + h cot ft when h > i' .-. A L sin 6 (which = A log sin 6) = log (1 + h cotO ) = M j(h cot - y) - i (h cot 6 - ^ V &c| = M h cot 6 - \ M h 2 cosec 2 (9 where M = /log e 10, and .*. < \, and powers of h higher than h 2 are neglected. When 6 is very small cosec 6 is very large, and the diff's are sensible but irregular. When 6 is nearly TV I2 cot 6 is very small, and cosec 6 nearly unity, so that then difTTs are insensible and irregular. But, with these exceptions, we have from above that A L sin 6 = M h cot 0, and .*. oc h So that the principle holds for L sin 6. h 2 § 57. cos (6 + h)/cos 6=1 h tan 6, when h ^> i' .-. A L cos 6 (which = A log sin 6) = log ji - (h tan 6 + j)l = - M j(htan0+ j)+ l(h tan + ^f&c} = — M h tan 6 sec 2 6 2 When 6 is very small tan 6 is very small, and sec 6 is nearly unity, so that then the diff's are insensible and irregular. 192 Trigonometry — Chapter X When is nearly irh, sec is very large, and then diff's are sensible but irregular. But, with these exceptions, we have from above A L cos = — M h tan 0, and .-. oc h .\ the principle holds for L cos § 58. tan (0 + h)/tan = (1 + ^^)/> -tan h tan 0) = (1 + h cot<9)/(i - h tan 0) /. A L tan (which = A log tan 0) = M log e (1 + h cot 0) — M log e (1 — h tan 6) *„(u a (hcotO) 2 } . . C, . (htantf) 2 ) = M j h cot# — L > + M jh tan 6 + '-> M h 2 = 2 M h cosec 2 6 + — (tan 2 - cot 2 <9) 2 x When # is small cot 6 is large, and the difFs irregular. When 6 is nearly 77-/2 tan 6 is large, and the diff's irregular. But, with these exceptions, we have A L tan 6 = 2 M h cosec 2 6, and .-. dc h /. the principle holds for L tan 6. § 59. The principle, for the remaining three T. F.s, can now •easily be deduced from the former three, by considering that L cosec 6 = 20 — L sin 6 L sec 6 — 20 — L cos 6 L cotan 6 = 20 — L tan 6 Or they can be deduced by putting for 6 its complement. .-., unless 6 is small or near 77/2, ify(#) is any T. F., and 8 an increment of 0, L/(<9 + <$)- L/(0) ocS Examples and Exercises 193 Examples 1. Find L cos 25 36' 19". In the table-book we find that L cos 25 37' = 9-9550653 diff' for i' = .0000606 .-. if L cos 25 36' 19" = 9-9550653 + x then x = -~ X .0000606 60 = -00000191 .'. the answer is 9-9550844 2. Given that L sin a = 9-6448213,^7^ (X. In the table-book we find that L sin 26 11' = 9-6446796 difF for 1/ = .0002569 .-. if a = 26°u' + x 7 ' then x = — -~ x 60 = 33 2569 ,\ a = 26°n / 33 // Exercises 1. Given that L sin 32 28' = 9-7298197, and Lsin 32 29' = 9.7300182, find L sin 32 28' 36". 2. Given that L tan 21 17' = 9-5905617, and L tan 21 18' = 9.590951, find L tan 2i°i7' 12". 3. Given that Lcot 72 15' = 9.5052819, and L cot 72 16' = 9-5048538, find L cot 7 2 15' 35". 4. Given that L cos 22 28' 20"= 9.9657025, L cos 22 28^ 10^= 9.96571 12, and L cos OC = 9-657056, find OC. 5. Given that L cos 20 35' 20"= 9.9713351, difference for 10" = -0000079, and L COS OC = 9-9713383, find OC. 6. Given that L cot 44°59' = 10-0002527, difference for i 7 = .0002527, and L cot OC = 10-0001234, find OC. o 194 Trigonometry — Chapter X 7. Show that, by taking logarithms of the particular case of Eulers formula when the angle is a right angle, viz. IT It 7T _ 2 cos — cos — cos — &c = — , 4 8 16 it retaining ten of the cosines, and using a seven figure table-book, we shall get 77 correct to five decimal places. § 60. If/ (0) denote any T. F.— A_/ (0) is irregular or insensible according to this table 6 A sin 6 A cos0 Atan0 Acot0 A sec 6 A cosec 6 o neither both neither irreg' both irreg / IT/2 both neither irreg 7 neither irreg r both A \-f(6) is irregular or insensible according to this table 6 ALsin0 A L cos 6 ALtanfl A Lcot0 A Lsec0 A L cosec 6 o irreg 7 both irreg 7 i rreg 7 both irreg 7 77/2 both irreg 7 irreg 7 irreg' irreg 7 both Note — In the above tables the o and 7T/2, under 6, are not intended to indi- cate that is absolutely o or 77/2 , but near those values. Hence when an A is small, it should not be found from its cosine, or cotangent, or cosecant; when it is near a right A, it should not be found from its sine, or tangent, or secant. Proportional Differences 195 We may however get round the difficulty by artifices. For if 06 is very small, so that COS 06 = a, where a is very nearly unity, then sin — = V(i — a)/2 (X from which — , and .*. 06, can be easily found. 2 Again if 06 is nearly a right A , so that sin (X = a, where a is nearly unity, then sin (45° — — ) = V(i — a)/2 06 from which 45 ■, and /. 06, can be found. 2 So also when 06 is nearly a right A , so that tan 06 = a, where a is very large, then tan (OL — 45 ) = a — 1 /a + 1 from which 06 — 45 , and .*. 06, can be found. Besides the above we have seen (in § 53) that the diff's are irregular for tan 06, when 06 > 77/4, and for cot 06, when 06 < irU ; so that these exceptional cases must be added to the above. From the 2nd table, we see that when an A is small, or near a right A, it cannot be accurately found from any one of its logarithmic T. F.s. § 61. As the case when 6 is small is one of frequent occurrence in practical work, various methods to overcome the difficulty of calculating L sin 0, L cos 0, and L tan 6, when 6 is small, have been devised. Such methods are of the greater value because in practice A s are more frequently reckoned from their loga- rithmic than from their natural T. F.s. o 2 i()6 Trigonometry — Chapter X Method i — For ordinary tables the following method (Maske- lynes) is available. Since sin = — ■<- 6 and cos = i 2 ) >■ for small A s , i- e. when > 3 sin# 2 / fry / '-0- = I -^ = \ I -V nearly = (cos 6)^ .-. log sin 6 = log 6 + ^ log cos 6 Now for small A s diff's of log COS 6 are insensible. .*. from this eq r n, given log 6 we can find log sin 6. The practical rule (given in Chambers' Tables, p. xviii) is "To the logarithm of the angle, reduced to seconds, add 4«68557 49, and from this sum subtract one-third of its L-secant, the index of the latter logarithm being previously diminished by 10: the remainder is the required L sin 0." The rationale of the Rule is easily seen to follow from the formula thus — If T = n" then = 7rn /(180 x 60 x 60) = 3-141592 x n /648000 .*. log#= log n + log 3-141592 — log 648000 Now, from table-book, log 3-1415 = -4971371 diff ' for 9 125 „ „ 2 28 .-. log 3-141592 = ■497 I 49 8 8 But log 648000 = 5-8115750 Subtracting, we get 6-68557488 Proportional Differences 197 .*. log# = log n + 6-6855749 .*. , by Maskelyne's formula, L sin = log n + 4-6855749 - £ (Lsec - 10) which is Chambers' Rule. Example To find Lsm i° 44' 36^.8 Here n = 6276^-8, and its log = 3'7977383 Constant number is 4-6855749 8.4833132 ^ (Lsec n — 10) = ^ x .0002011 = .0000670 .-. L sin i° 44' 36". 8 = 8.4832462 Conversely — given Lsin to find — for the given value of L sin find the nearest value of given in the tables, and use this to get L cos 6 ; which, since the diff's of L cos 6 for small A s are insensible, will be a sufficient approx': then, from the formula log 6 = log sin 9 — I log cos 6, we get a closer ap- prox' to 6. The practical rule (given in Chambers' Tables, p. xix) is "To the given L sin 6 add '5-3144251, and one-third of the Lsec#, the index of the latter logarithm being previously diminished by 10, and the sum will be the logarithm of the number of seconds in the angle ". To show that this Rule comes from the formula, we have, as before, log n = log 6 - 6-6855749 /., by Maskelyne's formula, = L sin + 5.3144251 + i (L sec - 10) which is Chambers' Rule. 198 Trigonometry — Chapter X Example To find 0, ifL sin 6 = 8-4832462 The 1st approx' to 6 is i°44' 36" and L sec i° 44' 36" = 10-000201 1 .\ -ij(Lsec0 — 10) = -0000670 L sin 6 = 8-4832462 constant number is 5-3144251 sum = 3-7977383 which, from tables, = log 6276-8 .-. n" = 6276 /, -8, where n" = r d = i°44' 3 6".8 For the L-tan gents, we get log tan 6 = log sin — log cos 6 = log — § log cos 6 Exercises Deduce, as above, the following rules, given in Chambers' Tables. 1. To find L tan 6, when 6 is small — " To the logarithm of Q, reduced to seconds, add 4-6855749, and two-thirds of L sec 6, the index of the latter logarithm being previously di??iinished by 10 : the sum is the required L tan 6." As a particular case find L tan i° 44' 36"-8. 2- If L tan 6 is given, 6 being small, to find 6 — " To the given L tan 6 add 5-3144251, and from this sum subtract two- thirds of the corresponding L sec 6, previously diminishing its index by 10 : then the remainder is the logarithm of the number of seconds in the angle 0." As a particular case find 6, if L tan 6 = 8-4834473. Method 2 — Delambre has given a method founded on the con- struction of a table giving the value of , sin . . „ log— 7j— + Lsin 1 for every second, for values of 6 from o to 3 . Proportional Differences 199 The mode of using this table is seen thus — If n" = T then 6 = n sin 1" (Cor' to Example 2, Chap / IX) . sin 6 , sin n" ••• log —n~ = log 6 n sin 1" = L sin n" — L sin 1" — log n .-. L sin n" = log n + ( log —w- + L sin 1" J which, by the special tables, gives L sin n" when 6 is known. Conversely — given Lsin n", to find n. First find an approx' value of 6, and then from the special tables get 1 sin 6 . . „ log — ^- + L sin 1" u Delambre's formula then gives log n, and n is known from the ordinary tables. Note — The error produced by using the approx' value of in log — 3 — , in- u stead of the true value, will be insensible for 7 dec'l places. Forif/(0) = log^ a ^vm 1 sin (0 + h) . sin A/(fl)-log- e + h -log-gr- r (0 + h) 2 ) . / 2 \ /0 2 0*\ r(0 + h) 2 (0 + hy } = \6 + Y2) ~ \~~6~ + 72 ) a u = + terms not affecting 7th place. 3 Method 3 — Finally the difficulty may be got over by a table giving values of L sin 6 for every second for values of 6 from o M h 2 to 3 : in such a table the term cosec 2 may be neglected. 2oo Trigonometry — Chapter X Exercises 1. Show that L cosx — L cos (x + h) h 2 # = (h tan x + — sec 2 x + &c)/log e 10. 2 . Prove the truth of the formula log n = Lsin n" + (10 - Lcosn")/3 — Lsin i" Note — This is an amalgamation of ' Delambre 's and Maskelyne' s formula. 3. An angle, which is known to be about two-thirds of a right angle, has to be found from its natural tangent, by means of a table-book which goes to seven decimal places : how nearly can the angle be calculated ? 4. Show that a large error may be expected in determining OC from the formula cot OC — tan OC = 2V i + cos 4 OC/Vi — cos 4 OC when OC is small, or near 45 , or near 90 . Also show that, if OC = 30 , and m is the error in the registered value of cos 4 OC, the value deduced for tan OC will be less than the correct value by 2V3m/9. ' Joh' CamV'. '37. 5. When an angle is nearly 64 36' show that it can be determined from its L sin within about one-tenth of a second, if the tables go to 7 decimal places, and log e 10 . tan 64 36' = 4.8492. Joh' Camb': '50. 6. In finding an angle (which is nearly a multiple of a right angle) from a table of log -sines calculated to n decimal places, show that the greatest error which can be committed, measured in seconds, is tan 6/{io n . 2 sin 1") very nearly. Joh* Camb 1 : '50. CHAPTER XI Relations between the Sides and Angles of a Triangle § 62. Def's — In the trigonometrical treatment of a triangle the following notation will be invariably used, unless the contrary is expressly stated — A, B, C for either the positions of the corners ; or the magnitudes of the angles at those corners : a, b, c for the sides respectively opposite A, B, C : S for the semi-perimeter ; so that 2S= a + b + c: A (the ordinary symbol for the word ' triangle '), when used alone, for the area of the triangle : s u s 2> s 3? respectively for s — a, s — b, s — - c; so that 2s 1 =b + c — a, 2 s 2 = c + a — b, 2 s 3 = a + b - c : D for the diameter of the circum-circle. Note — Si + s 2 + s 3 = s. § 63. Theorem — The sides of a triangle are respectively propor- tional to the sines of the respective opposite angles. Let ABC be a A; then of any two of its A 9 A, B, either both are acute, as in fig' (1) or one is acute, and one obtuse, as in fig' (2) or one is acute, and one right, as in fig / (3) 202 Trigonometry— Chapter XI .-., if p is the ± from C on AB, in all cases • A P P sinA = C* = B b ; and, by analogy, " sin A "~ sin B' ' c sin C Co/ (i) — A geometrical interpretation can be given to the ratios a/sin A, b/sin B, c/sin C. For, if D is the diam' of the circum-O, then, by a well-known geometrical theorem (see Euclid Revised, p. 288) D.p = ab D _ ab ___ ab a p ~ b sin A "~ sin A i. e. each of the above ratios = circum-diam / of ABC Cor* (2) — In any ratio, whose terms involve the sides of a A homogeneously, each side may be replaced by the sine of the A opposite ; and vice versd. For example, the ratios sin 3 A + sin 2 B sin C : sin A sin B sin C — sin A sin 2 C and a 3 + b 2 c : abc — ac 2 are equal, and .*. interchangeable. § 64-. Theorem — Any side of a triangle can be expressed in terms of the cosines of the angles ofivhich it is a common arm, and the remaining sides. c c c Relations of Sides and Angles 203 Take any side c of a A ABC : then the A s A, B, of which it is a common arm, may be either both acute, as in fig' (1) or one acute, and one obtuse, as in fig' (2) or one acute, and one right, as in fig' (3) .-., if CN is the X from C on AB, c = AN + BN, in fig 7 (1) = b cos A + a cos B c = AN - BN, in fig' (2) == b cos A — a cos (180 — B) = b cos A + a cos B c = AN, in fig' (3) = b cos A + a cos B, since then cos B = o, .-., in all cases, c = a cos B + b cos A and, by analogy, b = c cos A + a cos C a = b cos C + c cos B Cor 1 (1) — These results may be written — a + b cos C + c cos B = o\ acosC — b + ccosA= or acosB + bcosA — c =0 The eliminant of which, with respect to a, b, c, is — 1 cos C cos B cos C — 1 cos A cos B cos A — 1 which, expanded by Sarrus' rule, gives 2 cos 2 A + 2 n cos A = 1 a result already found otherwise (see p. 79). Cor' (2) — Again, the results may be written = o 204 Trigonometry — Chapter XI a — b cos C — c cos B = o ^ (b — c cos A) - a cos C — o . cos B = o (c — b cos A) — o . cos C — a cos B = o The eliminant of which with respect to — cos C, — cos B, is a b c (b — c cos A) a 0=0 (c — b cos A) o a Whence, at once, a 2 = b 2 + c 2 — 2 be cos A Note — This last result may also be got by multiplying the eq'ns (as originally found) by c, b, a respectively, and then adding the 1st two results so found, and subtracting the 3rd ; but, on account of its importance, an independent in- vestigation of it, on first principles, will be given. § 65. Problem — To express the sine of an angle of a triangle in terms of the sides of that triangle. A Let A, of A ABC, be acute in fig 7 (1), and obtuse in fig' (2) Drop CN ± on AB, or BA produced. Then, in fig' (1) a 2 = b 2 + C 2 — 2c. AN, by Euc ii. 13 and, in fig' (2) a 2 = b 2 + C 2 + 2 c . AN, by Euc ii. 12 .-., in both cases, 4 c 2 . AN 2 = (b 2 + c 2 - a 2 ) 2 Also AN 2 = b 2 - CN 2 .-. (2 c . CN) 2 = 4 b 2 c 2 - (b 2 + c 2 - a 2 ) 2 = (2 be + b 2 + c 2 - a 2 ) x (2 be - b 2 - c 2 + a 2 ) Relations of Sides and Angles 205 = (b + c + a) (b + c - a) (a - b + c) (a + b - c) .-. (2 c. b sin A) 2 = 16 s s x s 2 s 3 .-. sin A = — Vs SjSgSs In the case when A = 90 , a 2 = b 2 + c 2 , and this expression for sin A reduces to unity, as it should. Note — The expression Vs Sj s 2 s 3 is usually written Vs (s - a) (s - b) (s - c) ; but the shorter form is very convenient. § 66. Problem — To get expressions for the area of a triangle. Taking the figs 7 of the preceding §, we have A = ic . CN = \ be sin A Whence also A = v / ss 1 s 2 s 3 But 16 ss, s 2 s 3 = 2 b 2 c 2 + 2 c 2 a 2 + 2 a 2 b 2 - a 4 - b 4 - c 4 m m m A — 1 4 V2 2 b 2 c 2 - 2 a 4 Again A = 1 2 be sin A i a sin B asin C 2 sin A sin A a 5 sin B . sin C sin A 2 sin A § 67. Problem — To express the cosine of an angle of a triangle in terms of the sides of that triangle ; and thence to find the trigono- metrical functions of half that angle in terms of the sides. Taking the figs' of the preceding §, we have as before a 2 = b 2 + c 2 - 2 c . AN, in fig' (1) = b 2 + c 2 — 2 c . b cos A and a 2 = b 2 + C 2 + 2 c . AN, in fig' (2) = b 2 + c 2 — 2 c . b cos A 2o6 Trigonometry—Chapter XI .-., in both cases a 2 = b 2 + c 2 — 2 be cos A b 2 + c 2 - a 2 or cos A = 2 be When A = 90°, a 2 = b 2 + c 2 , and this expression for cos A reduces to zero, as it should. AAA Next, to deduce sin — , cos — , tan — , &c. 2 2 2 A Since 2 sin 2 — =1 — cos A 2 . 2 A 2 be - (b 2 + c 2 - a 2 ) .-. sin 2 — = ^-j 2 4 be - (a + b — c) (a - b + c) 4 be • f\ / Sq So or sin 2 / V be Also 2 cos 2 — = 1 + cos A 2 2 A 2 be + (b 2 + c 2 - a 2 ) 2 4 be (b + c + a) (b + c — a) 4 be A /ss! •• C0S 7 = V b^ By division tan — = 2 v sSj Whence, of course, sim'r expressions follow for the reciprocal functions. Relations of Sides and Angles 207 Note (1) — From these expressions sin — cos 22 be be .'. sin A = — Vs s x s 2 s 3 as before Note (2) — Of course the expressions for sin A and cos A, in terms of the sides, are deducible each from the other. § 68. Problem — To find formula connecting two sides of a triangle, and their included angle. sin B — sin C b — c Since sin B + sin C b + c B + C . B - C 2 cos sin 2 2 b — c .B + C B-C b + c 2 sin cos X B + C A B-C b-c \ cot tan 2 b + c B-C b-c A .-. tan = 7— — cot— . . (1) 2 b + c 2 v ' . . a sin A Again, since b — c sin B — sin C A A 2 sin — cos — /, x 2 2 '• a=(b - C) B + C . B-C 2 cos sin 2 2 • B -° /u \ A .-. a sin = (b — c) cos — (2) B — C A Sim'ly a cos = (b + c) sin — (3) Formulae (1) (2) (3) above give relations as req'd. 2o8 Trigonometry — Chapter XI N. B. In all the Examples and Exercises following it is to be understood that the notation defined in § 62 is adopted, unless some other is expressly in- dicated. Examples A+B - r -, pro Math' Tri': '51, A+B 1. If a tan A + b tan B = (a + b) tan , prove that a = b From given cond'n we get a (tan A - tan *±*) = b (tan *±* - tan b) . A-B . . A-B a sin b sin 2 2 A A+ B A+B _ cos A cos cos cos B 2 2 . A-B .*. either sin = o 2 sin A sin B or t- = b cos A cos B But either alternative gives A = B .-. a = b 2. Show that the sides and cotangents of the semi-angles of a triangle are so related that, if the one set are in arithmetical progression, so also are the other. Math' Tri'\ '64. A /ssj ss x v We have cot - - V — ^ = -r- = As i sa y 2 s 2 s 3 A B C Sim'ly cot - = X s 2 , and cot — - A s 3 But if a, b, c are in A. P. so also are s - a, s - b, s - c ; i. e. so also are s x , s 2 , s 3 ; A B C and .'. „ cot — , cot — , cot — 7 222 The converse is also evidently true. Examples 209 3. If tan A, tan B, tan C are in harmonical progression, show that a 2 , b 2 , c 2 are in arithmetical progression. From given cond / n we get that cos A cosC 2 cos B + sin A sin C sin B b 2 + c 2 - a 2 b 2 + a 2 - c 2 2 (a 2 + c 2 - b 2 ) 2abc 2 abc 2abc .-, 2 b 2 = 2 (a 2 + c 2 - b 2 ) .-. 2 b 2 = a 2 + c 2 i. e. a 2 , b 2 , c 2 are in A. P. 4. If ABC is a triangle, and x, y, z connected with its angles by the relations 2 (y - 2) cot- =0, 2 (y 2 - z 2 ) cot A = o prove that y 2 + z 2 — 2 yz cos A _ z 2 + x 2 — 2 zx cos B x 2 + y 2 — 2 xy cos C sin 2 A sin 2 B = sin 2 C Leudesdorf : E. T. xli. A A We have 2 (b — c) cot— oc 2 (sin B — sin C) cot — K 2 y 2 v . B-C A oc 2/ sin cos — 2 2 ^ . B-C . B+C oc2/Sin sin 2 2 oc 2 (cos C — cos B) and .'. = o Also 2 (b 2 - c 2 ) cot A oc 2 sin (B + C) sin (B - C) cot A oc 2 sin (B - C) cos (B + C) oc 2 (sin 2 B — sin 2 C) and = o .*. , if XYZ is the A whose sides are x, y, z, then A XYZ is sim'r to A ABC x 2 y 2 z 2 sin 2 A sin 2 B sin 2 C whence the req'd relations follow at once. 2IO Trigonometry— Chapter XI 5. Given the distances of each comer of a triangle PQR, from the sides of a known triangle ABC, to find the area of PQR. For brevity the fig' is left to explain itself : note that all the construction lines are J_s. LC = (/3 2 + 0C 2 cos C) cosec C MC = (/3 X + a x cos C) cosec C NC = (J3 3 + a 3 cos C) cosec C .-. LM = {/3 2 - jSi + (« 2 - «i) cos C} cosec C LN = {/3 2 - (3 S + (Og - QQ cos C} cosec C MN = {/3j - /3 3 + («i - a 3 ) cos C] cosec C Now 2 area PQR = (0C X + 0C 2 ) LM + (a x + (X 3 ) MN - (a 2 + a 3 ) LN and .-. = (a x + 0C 2 ) {/3 2 - ft + (a 2 - 0C ± ) cos C} cosec C + (a x + a 3 ) {ft - ft + («! - a 3 ) cos C} cosec C + (0L 2 + 0C 3 ) {ft - ft + (Og - a 2 ) cos C} cosec C = cosec C {(a 2 + a x ) (ft -ft) + C«i + a 3 ) (ft-ft) + (a s + a 2 ) (ft-ft)} = cosec C {a a (/3 2 - ft) + a 2 (ft - ft) + a 3 (ft - /3 2 )} Examples 2JI .-. area PQR = cosec C «i 0L 2 0C S ft ft ft cosec C 4ab A a^ a 0C 2 a a 3 bft b/3 2 b/3 3 2 A 2 A 2 A But aCXj + b^j + cyj = 2 A, and two sirr/r eq'ns. \ area PQR = abc - B) = cz/sin (^ - C) = abc/x sin + y sin (£ + z sin ^ Wolstenholme : 582. Brill: E. T. xlii. We have a 2 = y 2 + z 2 — 2 yz cos (1) b 2 = z 2 + x 2 — 2 zx coscf) (2) c 2 = x 2 + y 2 — 2 xy cos v (3) be sin A = yz sin 6 + zx sin (f) + xy sin)( .... (4) (2) + (3) - (0 gives be cos A = x 2 — xy cos x — zx cos

) Whence abc/2 (x sin 0) = ax/sin (6 — A) and sim'ly = by /sin ((f) — B) or = cz/sin (x — C) Note — This is a very useful theorem ; for, by means of it we can find x, y, z, when given 6, (j),X', or conversely. It is :. applicable for such p/ts as the centroid, orthocentre, Lemoine point. Stated as the Problem — Given a, b, c, 6, (j>, \, to find x, y, z — it is known as Hyparchus' Problem ; and is a familiar application of Trigonometry to practical surveying. P 2 2 _ c 2) (8) S- a 2 - b 2 8 sinj (A- B) sin B- C sin C-A ab sin 2 C 2 be sin A sin B sin C B — tan 2 — ta a) (c — a) 2 -?} = 1 (10) 2 A cos* *) = K A3 (± - ') \a 2/ abcs 1 s 2 s 3 \A D/ . v a cos 2 (B - C) n _ / (11) 2 ^ ^ = 82 (a cos A) cos B cos C v J (12) afbeos 2 c cos 2 -Y = (b - c) (b 2 2 C COS 2 — — C z COS' (13) 2 {as (b"2 + c*) cos A} = a2 b* c* 2 a^ (14) [ 2(a 2 -b 2 ) -i|_ abc Lcos 2 B — cos 2 AJ sin A sin B sin ( (15) D a a z cos" 1 — b b 2 cos 2 — 2 c c^ cos 2 = o Exercises 215 (16) 2 A 3 ♦ A + B * C cot — cot — cot — 2 2 2 * B + C * A cot — cot — cot — 2 2 2 cot — cot — cot — 2 2 2 3abc — 2 a 3 Note — All of this group are taken from the Oxford University First Public Examinations. 5. If is the orthocentre of a triangle ABC, prove that 2(a/OA) =]I(a/OA) 6. Prove that a 3 cos 3 B + b 3 cos 3 A = c 3 — 3 abc cos (A — B) Ox? Juri Math' Schol': '89. 7. If A, B, C are in arithmetical progression, prove that 2 cos = a 3 + c 3 /b 3 = a + c/v a 2 — ac + c 2 8. If a, b, C are in harmonical progression, prove that B cos — = Vsin A sin C/(cos A + cos C) 9. A triangle is formed by a parallel through A to BC, and perpendiculars at B, C to BA, C A respectively ; show that its area is a 2 cos 2 (B - 0)1 2 sin A cos B cos C 10. If a is the length of the line bisecting a triangle, and making equal angles with the sides terminating it ; /3, y similar lengths ; prove that 2(a 2 /3 2 ) =4 A 2 ^ Q.C.B. '69. 11. If d is the distance between the centres of two circles whose radii are R, r; show that the distance between the two points on their line of centres from which they appear equally large is 2 d R r/ (R 2 - r 2 ). 12. Show also that they appear equally large from all points on the circle whose diameter is the join of these points. (See Euclid Revised, p. 344, Ex' 6) 216 Trigonometry — Chapter XI 13. The sides of a triangle are 141, 100, 53 feet; find the length of the perpendicular on the longest side from the opposite corner. 14. In the sides BC, CA, AB respectively of a triangle are taken points P, Q, R, so that BP/a = CQ/b = AR/c = A show that 2 PQ 2 = (1 - 3 A + 3 ^ 2 ) 2a 2 15. If C = 6o°, prove that 1 1 + a+c b+c a+b+c R. U. SchoV: '82. 16. If the angles of a triangle are in arithmetical progression, and 5 is their common difference, prove that tan 2 5 = 4 b 2 - (a + c) 2 /(a + c) 2 R. U. SchoV-. '84. 17. Show that the equations 2 (ax yx 2 — a 2 ) = 2 abc 2 V(x 2 - a 2 ) (x 2 - b 2 ) (x 2 - c 2 ) = x (a 2 + b 2 + c 2 - 2 x 2 ) have a common root D. 18. If cos 6 = a/b + c, cos 1) and on the lesser of them as base a similar triangle is described having again OL as vertical angle : this process is continued for ever : show that the sum of all the triangles is (1 H s 1 original A. m 2 — 2 m cos OC/ b Pet' CamV: '66. Exercises 217 21. Prove that 2 V(4 b 2 c 2 - A 2 ) (4 c 2 a 2 - A 2 ) = A 2 22. The centres of three circles, which touch each other externally, form a triangle whose sides are 3, 4, 5 inches : prove that the area included between the circles is slightly more than half a square inch. Assume that sin - — = -8 10 23. If sin A, sin B, sin C are in harmonical progression, prove that so also are 1 — cos A, 1 — cos B, 1 — cos C. 24. If cos A, cos B, cos C are in arithmetical progression, prove that s x , s 2 , S 3 are in harmonical progression. 25. A pyramid is cut off from a cube; if OC, j3, y are the distances of the corner of the cube from the corners of the plane of section, show that the area of the section is VS (OC 2 /3 2 ) /2. 26. If P, P' are points in side BC ; Q, Q' points in side CA; R, R' points in side AB ; of a triangle ABC ; prove that, if perpendiculars to the sides at P, Q, R concur ; and perpendiculars to the sides at P r , Of, R' concur ; then 2 (PP' sin A) = o; where PP' is to be considered positive when in cyclic order ABC. Q. C. B. '67. 27. If D is the point in side AB of triangle ABC, such that CA : CB = DB : DA and if (f) denotes the angle BCD ; prove that C C (a 2 + b 2 ) cot - + (a 2 - b 2 ) tan - = 2 b 2 cot Then -. = tan 2 -*- b + c 2 So that the formula of § 68 may be written B - C . _.A , A tan = tan 2 -f- cot — 2 2 2 which is in a form adapted to log's The result may be written as B - C = 2 tan- 1 5 tan 2 i (cos- 1 cot -1 2. To adapt the first formula of % 67 to logarithmic calculation. The formula gives a 2 = V {(b + c) 2 - 2 be (1 + cos A)} + C ) Vi - 4 be cos 2 -/(b + c) 5 Cb + c) Logarithmic Adaptation 223 Now, since 4 be cos 2 — < (b + c) 2 , there must be some si n -1 ^2 Vbc cos — /(b + c)( less than 180 , call it 0, so that — A / sin 9 = 2 Vbc cos — / (b + c), where 6 is given by L sin0 = logb + logo — log(b + c) + Lcos — A 2 So that the first formula of § 67 may be written a = (b + c) cos0 which is in a form adapted to log's. The result may be written as a = (b + c) cos sin- 1 12 \/bc~cos — /(b + c) I Another modification of this formula may be got thus — a = V{(b - c) 2 + 2 be (1 - cos A) } = (b-c) V 1 +4bcsin 2 ^/(b-c) 2 Now there must be some tan- 1 -j 2 Vbc sin — / (b - c) I less than 180 , call it )(, so that A / tan x = 2 Vbc sin — / (b — c) where )( is given by L tan x. = log b + log c — log (b — c) + L sin — So that the first formula of § 67 may be written a = (b — c)sec)( which is in a form adapted to log's. The result may be written as a = (b - c) sec tan -1 12 Vbc sin— / (b — c)> 224 Trigonometry— Chapter XII A third modification of this formula may be made ; for a 2 = (b 2 + c 2 ) ( cos 2 ^ + sin 2 *) - 2 be ( cos 2 * - sin 2 -) = (b - c) 2 cos 2 - + (b + c) 2 sin 2 - 22 .-. a = (b - c) cos — v 1 + 1 -, tan - 1 v 2 \b — c 2/ (b + c A\ , tan — ) less than 180° call it w, b — c 2/ so that b + c . A tan co = : tan — b — c 2 where a) is given by L tan a) = log (b + c) - log (b - c) + L tan — So that the first formula of § 67 may be written a = (b — c) cos — sec w ' 2 which is in a form adapted to log's. The result may be written as a = (b - c) cos - sec tan -1 ( r tan — J v 2 \b - c 2/ Exercises 1 . Adapt to the use of logarithms — (1) sec (X — sec /3 (2) cos A - cos (A + a) - cos (A + /3) + cos (A + OL + fi) (3) tan = a sin (p/(.b - a cos b ; and, 2 , when a < b. 2. If ABC is a triangle, prove that a = (a + b) cos 2 2- 2 - b= (a+ b)sin 2 * 2 ' 2 Vss 3 C where sin C 9 , tan A = p-J- tan — cos 2 d> a — b 2 ^ ^ 2 b C „ . tan B = tan — cos 2 (b a — b 2 ^ /a+ b . where b, a' > b', a" > b", &c 14. Show that cos (a + x) = cos (X sin x + sin (3 may be reduced to sin/3 , cos (

^ sin a ^ , sin (45 + a) tan (b = ^-4 7Z ^ cos 45 cos OC 15. Adapt to logarithmic computation sin 2 x— 2sinx siny cos (x + y) + sin 2 y Catalan : E. T. 1891. CHAPTER XIII Solution of Triangles § 70. Def f — The Problem to be solved is this — Given three parts of a triangle (not the three angles) to find the other parts. Or we may put it thus — Given enough (and only enough) about a triangle to distinguish it from all other triangles, it will be said to be ' solved ' when its remaining details are determined. On consideration of the Problem in particular cases it is readily seen that its solution — i°, for right-angled A s is simple and immediate ; and, 2°, for A 8 not right-angled has to be effected by reducing such to right-angled cases. Note — What is called ' the solution of a triangle' is, " in truth, a reduction of it to the solution of a right-angled triangle ; and the maker of the tables it is who solves the right-angled triangle." (De Morgan) We have .*. to consider two classes of A s — those which are right-angled ; and those which are not right-angled — and there will be various cases in each class. Class I — When a A ABC is right-angled — say at C. i°, given a and b. tan A = a/b, whence A is known then B = 90 — A and c = a /sin A, or = b/sin B 2°, given b and c. sin B = b/c, whence B is known then A = 90 — B Also a = \/(c + b) (c — b) 9 2 228 Trigonometry — Chapter XIII 3°, given c and A. B = 9 o° - A b = c cos A a = c sin A 4°, given a and A. B = 9 o° - A b = a cot A c = a/sin A, or = b/sin B 5°, given b and A. B = 9 o c - A a = b tan A c = b/cos A, or = a /sin A Exercises Verify each of the above methods, by using a table book, in the case of the following triangle. A = 59 o'2i".2 5 , B = 3o 59' 3 8"-75, a = 110.0951, b = 66-1364, c = 128-4327 Note — 77/£ verification will be accomplished by assuming as data the two parts, corresponding to those of each the 5 cases, and finding the rem'g three parts as shown above. Class II — When a A ABC is not right-angled. i°, given the three sides a, b, C. We may use any of the formulae such as A /s,s ? A /sSj A /s 2 s 3 sin — = a / -^-j cos — = a / t— ^ tan — = 2 / V be 2 / V be " 2 / V ss x A A iVfrfc — If only one A is wanted, then by using sin — or cos — we shall 2} 2 only need two of the expressions s, s x , s 2 , S 3 . Also if — is between 45 and Solution of Triangles 229 A A 1 35° it is better to use cos — , but otherwise sin - : the reason for this choice 2 2 depending on the principle of the irregularity and insensibility of differences , summarised in § 60. A A R O If two A s are wanted use tan — : in this case if - , — , — are thus calcu- 2 222 lated, then the relation A^2 + B/2 + C/2 =^ 90 will give a verification. Or we may use such formulae as cos A = (b 2 + c 2 - a 2 )/( 2 be) and, if the aid of logarithms is desired, we have the adaptation effected in § 69, Example 2. Note — The last formula will do well without logarithms when a, b, c are small. Or again the solution may be effected thus — c c c Drop CN _L from C on AB, produced if necessary. Now AN 2 ~ BN 2 = b 2 ^ a 2 ' and AN + BN = c Whence AN + BN = b * " ^ c the upper signs going together, and the lower together. Hence AN and BN are known. Then cos A = ± AN/b, and cos B = ± BN/a the neg' sign being taken when the A is obtuse. These give A and B 5 and then C = 180 - (A + B) ^3° Trigonometry — Chapter XIII 2°, given two sides and the included A — say b, C, A. If all the other parts are wanted, we have, by § 68, B -v C , (b ^ c A tan -1 h cot — + C 2 II A B + C A and = oo 2 2 whence B and C are known. Then a = b sin A /sin B, or = c sin A /sin C But if a only is wanted, we have, by § 67, a 2 = b 2 + c 2 — 2 be cos A Note — The adaptations of the above formulee to logarithmic use are given in § 62, Examples 1 and 2 : the last will do as it stands if the numbers are small. Or the solution may be effected thus — G C Drop CN ± on AB, produced if necessary. Then CN = b sin A AN = + b cos A BN = c + AN = c + b cos A _ b sin A .-. tan B = — — c + b cos A a = (c + b cos A) sec B In addition to the foregoing — which are the methods usually given in this case — the following compact solution is worth notice. Solution of Triangles 231 Given b, C, A ; and that b > C. b sin B sin (A + C) , . , . AX . A - = — — ~ = \ — ^ — - = (cot C + cot A) sin A c sin C sin C v ' .-. cot C = — -. — x- — cot A c sin A b ( c cos A) " c sin A( b j = —I ~ (1 - cos 2 0) c sin A v ' x - b sin 2 6 . o /j c cos A 1. e. cot C = ; — t- > where cos u = c sin A' b Whence C is known, and then B = 180 — (A + C) and a = b sin A /sin B, or = c sin A /sin C 3°j given two A s and a side — say A, B, c. C = 180 - (A + B) a = c sin A/s b = c sin B /s or = c sin B /s nC n C n (A + B), if C is obtuse. 4 , given two sides and an A opposite one of them — say a, b and A — usually called ' the ambiguous case! If a sin B > b, then sin A > 1, and .-. there is no solution : i. e. the parts will not form a A. If a sin B = b, then sin A = i, and A = 90 ; so that c = a cos B, and C = 90 — A, complete the solution. If a sin B < b, then sin A < 1, and there are two solutions, one an acute and the other an obtuse-angled A- Suppose Bj, C x , c 1 the parts of the acute-angled A ; and B 2 , C 2 , c 2 „ „ obtuse „ 2 3 2 Trigonometry — Chapter XIII Then sin B x Q, c, b sin A /a 180 - (A + B x ) a sin Cj/sin A is a set of solutions And sin B 2 = b sin A /a C 2 = i8o°-(A + B 2 ) c 2 = a sin C 2 /sin A is another set of solutions .\ the preceding solutions belong, each of them, to one or other of the A s (a, b, B x ) (a, b, B 2 ) where B 13 B 2 are acute and obtuse supplements ; but both 3 1 and B 2 come from the eq'n sin B = b sin A /a. .•. the solution (as often occurs in Algebra) gives the solution of a cognate Problem — viz. for the A which has an A the supp't of the given A . Respecting the double solution, and the solution covering the cognate case, we notice that — If b < a, so that B < A, there is no obtuse value of B, so that both solutions belong to acute values of B. If b = a, so that B = A, and A, B are both acute, there is one solution — viz. an isosceles A — and the other degenerates into a straight line. If b > a, so that B > A, there is no obtuse value of A, but either value of B will do, and one solution belongs to each value. The cases when b > a, and b < a, are here illustrated. Solution of Triangles *33 The double solution indicates a quadratic : thus from the eq'n c 2 + b 2 — 2 cb cos A = a 2 we get Cj = b cos A + Va 2 — b 2 sin 2 A and c 2 = b cos A — Va 2 — b 2 sin 2 A The ' ambiguous case ' may also be treated geometrically as follows — A N Drop CN ± on AB. Then in fig 7 (i) — the ordinary 'ambiguous case' — AN = b cos A CN = b sin A B 1 N = Va 2 - b 2 sin 2 A = B 2 N_ ... Ci = AN + B, N = b cos A + Va 2 - b 2 sin 2 A ) c 2 = AN - B 2 N = b cos A - Va 2 - b 2 sin 2 A I And in fig' (2) — the ' cognate case ' — Va 2 - b a sin 2 A + b cos A | Cl = B 1 N - NA B: c 2 = B 2 N + N A = 7a 2 - b 2 sin 2 A - b cos A ) Note — Though all questions involving the solution of A s are reducible to the preceding cases, the data may be in forms not exactly corresponding to any one of them : thus among the data may be the sum of two sides or an altitude, or the area. The necessary modifications in the solution will readily suggest themselves. 234 Trigonometry— Chapter XIII Examples 1. If a/4 = b/5 = c/6, find B : log 2 is supposed known ; and L cos 27 53' = 9.9464040, L cos 27 54' = 9-9463371. We have cos * = x/ * 5 * 5 = V^ 2 4x6x4 32 B 7 .-. L cos 10 = 1 — - log 2 2 2 '/. L cos — = 9-94 6 395° = Lcos 27°53' + 8" .-. cliff' for 8" = -0000090 ,, „ 60" = .0000669 .-. 8" = #- x 60" = 8", nearly 669 .'• f = 27° 53' 8" .-. B = 55° 4^ 16" Note — Recollect that log 2 = .3010300, is given by 2 C sin (/>=+ — V(m + c) (m - c) sec Exercises 237 15. In the 'ambiguous case,' where a, b, A are the given parts (b being greater than a, and the double values of the other parts being denoted by suffixes) prove each of the following — (1) cot Cl + ° 2 = tan A 2 , „ sin Ct sin C 9 (2) - — =- x + . * = 2 cos A v sin B x sin B 2 (3) Distance between centres of circum-circles of the two triangles = (c x m* c 2 )/(2 sin A) , . 2 b + c, + Co 2 b — Ci — c 2 (4) L — T 1 + ^—k 1 = 4 b T 1 + cos A 1 — cos A * , . A x A 2 2 (a 2 + b 2 cos 2 A) (5) -r 1 + 2 _ o2 A 2 A x b (6) A x 2 + A 2 2 - 2 A x A 2 cos 2 A = p (A! + A 2 ) 2 (7) c i 2 - 2 Cj c 2 cos 2 A +■ c 2 2 = 4 a 2 cos 2 A (8) If A 1 = kA 2 , b/a lies between 1 and k + i /k - 1 16. In a triangle ABC, if AB = AC + \ BC, and P is the point in BC for which PC = 3 PB, prove that angle APC is half angle ACP. Math' Tri'\ '78. 17- Show that the three sides, area, and an altitude of a triangle, can always be represented respectively by the rational formulae (ux) 2 + (vy) 2 , (uy) 2 + (vx) 2 , (u 2 + v 2 ) (x 2 + y 2 ), uvxy (u 2 + v 2 ) (x 2 — y 2 ), 2 uvxy Dr. Hart: E. T. xxiii. 18. Given the base c, the altitude h, and the difference of the base angles (X ; find the other sides, and the vertical angle. If 6 1 , 6 2 are two values for this vertical angle, prove that cot 6 t + cot 2 = 4 h/(c sin 2 OC) Prove that only one of these is a proper solution ; and that if it is 6 1 , then tan — = ( V4 h 2 + c 2 sin 2 (X — 2 h)/c (1 - cos a) Account for the appearance of 0% . Wolstenholme : 546. 23 8 Trigonometry— Chapter XIII § 71. If in the use of any one of the preceding methods for a particular case, it should happen — as it sometimes does in prac- tical applications — that the ' differences ' are ' irregular ' or ' insen- sible/ then the method will not give accurate results. Such cases may sometimes be treated by considering that — if A is very small, i. e. < i', then its radian measure (0 say) < -0003 Now as sin A lies between 6 and 6 — @ 3 /6 and cos A „ „ 1 „ 1 — 2 fe .-. the error, by putting 6 for sin A, and 1 for cos A, will not affect the 7th dec'l place, and .\ will give results as true as can be got by an ordinary table-book. Examples 1 . If C = 90 , and b, A are given, where A is small, find c approximately . Here the formula c = b/cos A is not good, v (see § 59) the diff's for cosines of small A s are ' irregular.' , 1 — cos A . • o A But then c — b = b r — = 2 b Sin 2 — , nearly cos A 2 .\ c = b ( 1 + 2 sin 2 — V nearly 2. If a, b, C are given ; and also that C = 180 — k°, where k is very email, find an approximation for c. i_N C 4 ab . o kH By Example 2, p. 222, c = (a + b) )i - + 2 sin 2 -V = (a + b) | I -(^ry 2Sin2 ^' nearly 2 ab . „ k = a + b r sin 2 - a + b 2 ab -oi- k *= a + b pr sin^ k, since cos -= 1 2 (a + b) 2 ab = a + b - 2 a ib /TTkV + b)Vi8o/ if k ^> 10", see p. 179, Example 2, Cor' Solution of Triangles 239 The same substitutions may be used to find connections between small variations in sides and A s of a A, consequent on an error in the estimation of the size of one of them. Note — In what follows the letter 8 will stand for the words ' a small incre- ment of : thus 8 a means a small increment of the side a. All increments of angle are supposed to be in radian measure. Recollect that if A, B, C are A s of a A, then 8 A + 8 B + 5C=o, always. Examples 1 . In a triangle right-angled at C, suppose that while b is true, either a or A has a small error, producing a corresponding error in the other: required the connection between the errors. a = b tan A .*. a + 8 a = b tan (A + 8 A) sin b A ba= b cos A cos (A + 8 A) .\ b a = bsec 2 A . b A, approximately. 2. Given b, C, and A, and that while b and c are true there is a small error in A : required to find the corj'esponding error in B. h h sin B = - sin C = - sin (A + B) c c .-. sin (B + b B) = - sin (A + b A + B + 8B) c .-. sin B + cosB .8B = -{sin (A + B) + (8A + 8B)cos (A + BU .-. cos B. SB = --(8 A + 8B) cosC c .-. 8 B (cos B + - cos C J = 8 A cos C * ~ / r^ sin B \ sin B cos C. .-. 8 B ( cos B + -7—^ cos C ) = . ^ — 8 A \ sin C / sin C _sinBcosC gA sin A 240 Trigonometry — Chapter XIII Exercises 1 . Given A, C, c ; and that B = A + n", where n is small ; show that b = — — ^ (sin A + — — — - cos A J sin C \ 206265 / 2. If C is very obtuse, show that the number of seconds in A + B is nearly 206265 v 2 (a + b) (a + b — c)/(ab) 3. If the sides and angles of a triangle ABC receive small increments, show that — (1) when a and b are constant, c . b B + b cos A . b C = o (2) when a and A are constant, cos C . b b + cos B . § c =0 (3) when a and B are constant, tanA.8b-b.8C = o Joh' Camb'\ ' 43. 4. The angles A, B of a triangle have small unknown errors, show that sin (A + B) will have its least error when C is nearly a right angle. 5. The side b of a triangle is rightly known, but the angle B has an error whose cube may be neglected, show that the circum-diameter is b cosec B 5 1 - b B cot B + 2S_ ( co t 2 B + cosec 2 B)l PemV Camb'\ '85. 6. Solve a triangle when given the lengths of its three altitudes p l5 p 2 , p 3 ; and show that a small error in the measurement of p x will cause errors sin A. § p x in the value of A and — cot B cot C in the value of a p x sin B sin C h Pi Math' Tri f : '66. 7. If the sides of a triangle are slightly changed, prove that b D = 11 cot A . 2 (sec A . b a) CHAPTER XIV Lines and Circles remarkably associated with the Triangle § 72- Def's — With respect to a triangle ABC, and in addition to the notation fixed in § 62, the following lettering will be in- variably used, unless the contrary is expressly stated — I for the position of the in-centre : E x ,, „ ex-centre within the angle A F R t -2 33 33 33 33 » D F C *— 3 33 33 33 33 33 ^ S ,, ,, circum-centre'. O ,, „ ortho-centre: G ,, „ centroid'. K „ ,, *Lemoine (or symmediari) point \ N „ „ nine-point-centre : O ,, „ positive * Br ocard point: £2' „ „ negative *Brocard point'. r for the length of the in-radius : r x „ „ ex-radius corresponding to E 3 : r E • ■2 33 " 33 33 " 2 * r E • 1 3 33 33 'J 33 33 3 ' R „ „ circum-radius : Pi 3 P25 P3 >j 33 altitudes of A, B, C, respectively : O) for the magnitude of the *Brocard angle. * See Section ix of the General Addenda for the 3rd edition of Euclid Revised. This Section is published separately by the Clarendon Press (price 6d.) under the following title — An Introduction {within the limits of Euclidian Geometry) to the Lemoine and Brocard Points, Lines, and Circles. R 242 Trigonometry— Chapter XIV The results now to be given — though many of them are mathe- matically of great importance — do not involve any fresh principles, but are rather amplifications of the principles already given in Chapter XI. They will therefore be considered as Examples and Exercises, and printed accordingly. 8 73. Properties of some Lines of importance in connection with a known triangle. Examples 1. Lengths of the internal and external bisectors of an angle of a triangle. Let AX, AY be the respec- tive intern' and extern' bisectors A of BAC ; where X is in BC, and Y in BC produced. Then, since area AXB + area AXC = A and area AYB ^ area AYC = A .-. , if x = AX, and y = AY, we have \ ex sin — + \ bx sin — = \ be sin A and \ cy sin (90°+ — \ <+. \ by sin (90 - - J = J be sin A 2 be cos whence x = 2 _ 2 Vbc ssj b + c b + c 2 be sin — and y = 2 2 \/bc s 2 s 3 b ^ c Cor* (1)— XY 2 = x 2 + y 2 = (b^^{ (b - c)2cos2 i + (b + c)2sin2 7S Remarkable Lines and Circles 2 43 A b 2 c 2 -. XY 2 = -4, (b 2 + c 2 - 2 be cos A) XY = (b 2 ^ c 2 ) 2 2abc b 2 ^c 2 . A sin — 2 sin (90°+ — ) Cor* (2)— BX=x-r-^-= r^ - , and B Y = y ^ 2 -l K J sin B b + c y sin B ac ac b <** c 2. Lengths of the median, and symmedian, /r then, since 6 = 180 — A in above, we have . PX.QY 2 the cross ratio ww —- = cos- 5 — XY . PQ 2 5. Any line from A meets BC in P : if PB, PC, and the angles APC, PAB, PAC, are denoted respectively by m ; n, 6, (3, y; then m cot (3 — n cot y = (m + n) cot 6 = n cot B — m cot C AP sin (6 - j3) For — = m sin/3 AP sin (6 + y) and — = D sin y = (cot /3 — cot 6) sin d = (cot y + cot 6) sin 6 , m s\n(6 - B) Q . Q Also -r^ = ^ — = — - = (cot B - cot^) sin 6 and AP sin B n sin (0 + C) AP sin C whence above results follow at once (cotC + cot0) sin0 Cor' — If AP is a median, 2 cot 6 = cot B — cot C .*. 2 cot 6 — o, for the three medians 24# Trigonometry— Chapter XIV 6. Parts of a triangle in terms of its altitudes. Since 2 A = p x a = p 2 b = p 3 c all formulae, involving a, b, c homogeneously, will be true when for a, b, c are respectively put the reciprocals of p x , p 2 , p 3 . Hence come such results as tan 2 — 2 VPs Pi P2/ VPi P2 P3/ \Pi P2 Ps/ VP2 Ps Pi/ As an Exercise the Student should express A, s, S x , sin 2 — , &c. in terms of these altitudes. Exercises 1. AD is the bisector of A ; and P, Q points in BC, on opposite sides of D, such that the angles PAD, QAD are each equal to 6 ; prove that 1 1 1 1 ~ A AP + AQ : V + ^ = ° 0Sf > ;C0S T 2. If A x is the area of a triangle of which two sides and the included angle are respectively the bisectors of A, and A; and if A 2 , A 3 are similarly con- structed with respect to B, C ; prove that 111 ^7 + a7 = a7 where a, b, C are in order of magnitude. 3. If the bisector x, of the angle A, divides BC into parts m, n (m lying next B) prove that — (1) 2 Ra x sin — = m 2 b + n 2 c 2 (2) p x = mn (m + n) sin A/(m 2 + n 2 — 2 mn cos A) Also, if x makes an acute angle (j) with BC, and c is greater than b, prove that . , c + b A tan © = r tan — ^ c — b 2 4. If x lt x 2 , x 3 are the internal bisectors of A, B, C respectively ; prove that— (1) n{ Xl (b + c)/a} = 8s A (2) 2 {x x 2 (b + c) 2 /(bc)} = 4 s 2 Exercises 249 5. Prove that the area of the triangle formed by joining the points where the internal bisectors of the angles of a triangle meet the opposite sides is 2AlI{a/(a + b)} 6. If two internal bisectors of the angles of a triangle are equal show (by trigonometry) that the triangle is isosceles. Note — A simple geometrical proof of this will be found in E. R. p. 413. 7. If I, m, n are the medians of a triangle, prove that 3 A = V 2 2m 2 n 2 - 2 I 4 8. If the median from A divides the angle A into parts 0, (f), prove that — 0* cot cf) = cot B -w cot C 9. If C is a right angle, and (f) the angle between the bisector of A and the median from A, prove that tan d> = tan 3 — r 2 10. If the median from A = b = c/2, prove that sin 2 C = 4 and sin 2 B = -^ 11. BY, CZ are perpendiculars from B, C on the opposite sides, and AY, BZ are equal ; show that the cross of the median from B with CZ, and the cross of the median from C with BY, are on the bisector of A. 12. If m is the median from A, prove that {m 2 - (a/2) 2 } tan A = 2 A Hence prove that, if P is any point on a circle inscribed in a square ; and OC, (3 the angles subtended at P by the diagonals — tan 2 a + tan 2 /3 = 8 13. If a l5 b l5 Cj are the sides of the pedal triangle respectively next A, B, C, prove that a x b 2 + b x a 2 = b x c 2 + c : b 2 = c x a 2 + a x c 2 14. A new triangle PQR is formed by drawing tangents to the circum- circle at A, B, C : if XYZ is the pedal triangle of ABC, prove that area XYZ : area PQR = 4 IT cos 2 A : 1 250 Trigonometry — Chapter XIV 15. In Example 4 (2) show that the other five cross ratios of the range PXQY are expressible as the other five T. F.s of 6/2. Phil' Trans': 1871. 16. If x, y, z are the lengths of perpendiculars from A, B, C on any line, prove that 2 (a 2 x 2 ) - 2 2 (be yz cos A) = 4 A 2 Ox? 1st Public Exam'\ '77. Show that this theorem is equivalent to the following — "With A, B, C as centres; and p 1} p 2 , p s as radii; circles are described : the condition that these circles may have a common tangent is 2(^-22^(^ = 1 \Pi/ \P2P3 / Editor s Question : E. T. xx. 17. If I, m, n are the medians, prove that — (1) 2 {asinB V4 I 2 - a 2 sin 2 B + a sin C V4I 2 - a 2 sin 2 C} = 18 A (2) 2 [2asinBv / 4m 2 - b 2 sin 2 A} + 2 (a 2 sin 2 C) = 24 A M c Kenzie : E. T. xxvii. 18. If the bisectors of A, B, C meet the opposite sides in D, E, F respect- ively; and if x, y, z are the respective perpendiculars from A, B, C on EF, FD, DE; prove that s(-^-Y= 11 +8llsin^- Nash : E. T. xlix. D Q Prove also that tan EDF = tan 6 cos 2 where 2 tan — = sec — 2 2 /. /. Walker : E. T. xxiv. 19. Taking the figure of Example 2, if (p denote the angle APC, and )( the angle AMC, prove that — (1) cos^)/cosx = cos A (2) tan j- — ^ = r- tan — v 2 c + b 2 j. c 2 + b 2 x . (3) tan (j) = q2 _ b2 tan A /. /. Walker : E. T. xiii. Remarkable Lines and Circles 251 20. Taking the definition of, and the theorems about, the mean centre of any number of points, given on pages no to 112 of Euclid Revised, show that the mean centre of points A, B, C — (1) for multiples sin A, sin B, sin C is I : (2) „ —sin A, sin B, sin C is E x : (3) (4) (5) (6) sin 2 A, sin 2 B, sin 2 C is S tan A, tan B, tan C is O : sin 2 A, sin 2 B, sin 2 C is K sin 2 B + sin 2 C, &c. is N. 21. If M is the mean centre of A, B, C for multiples I, m, n ; and if MA = x, MB = y, MC = z, and 2 a = Ix + my + nz ; prove that area ABC = (I + m + n) W {a — \x) (a — my) {a — nz)Al m n) Brill: E. T. xlviii. § 74-. Properties of the circum-, in-, and ex-circles. Examples 1. Length of the circum- radius. A Let A of A ABC be acute in fig' (1), and obtuse in fig' (2). A Draw the diam r BD of the O round ABC ; then BCD is right. A A A Now BDC = A in fig r (1) ; but = 180 - A in fig r (2 ) .'. , always, a — 2 R sin BDC' = 2 R sin A .-. abc = 2 R be sin A = 4 R A abc abc or R 4 A 2 sin A 2 sin B 2 sin C 25^ Trigonometry— Chapter XIV 2. Length of the in-radius. By Eu 2 bc- = r~^R (13) 2 (p 2 + ps)/^ = 6 (14) r 2 r 2 + 2 r 2 r 3 = 2 ab /A B C \ (15) r = 2 R (cos 2 sin 2 sin 2 — ) V 2 2 2 / * A u + B * c a tan — b tan — c tan — (16) ■ — - = ■ r — n r- r 2 r - r 3 (17) 2a 2 = 2s 2 - 2r 2 - 8Rr (18) 2ab = s 2 + r 2 + 4Rr NOT'E-^Tke last two results will come by elim'g 2ab and 2 a 2 successively between 2 a = 2 s, and s r 2 = s x s 2 s 3 . (19) lEj/sin — = IE 2 /sin — = IE 3 /sin — (20) IEj = 2 VR(r! - r) (21) r.lE 1 .IE 2 .IE 3 = (abc/s) 2 (22) AI.AE 1 = bc (23) 2Ja 2 .IE 1 2 ^b 2 -c 2 ) cosec 2 ^| = (24) n E x E 2 = 8 II (r x cosec A) (25) (4 R + r) (4 R + r + s VI) (4 R + r - s VI) = 2 rV - Ox* 1st Public Exam': '90. 2 iV - 3 i"! r 2 r 3 258 Trigonometry— Chapter XIV (26) If (r 2 + r 3 - rO (r 3 + r x - r 2 ) (r x + r 2 - r 3 ) + 8 r x r 2 r 3 = o then r + 4 R = 2 s Ox' Jun' SchoV: '79. (27) Vabc 2 \/— = 16 R Vr II cos ~ r i 4 Oxf Jun' SchoV: '74. < ox * locr 2 ( S ' n B - sin C ) (28) tan ISE, = — r y 1 2 cos A — 1 Leudesdorf'. E. T. xxxii. (29) 2 (a 2 cot^) : 2 (a 2 tan ^) = R + r : R - r (3 o) 3 V^-V^-V^-\ /F P-s (31) a(s 2 -AP 2 ) = 4 ss 2 s 3 where P is point in which ex-circle E. x touches BC. Math' Tri': '66. 2. If Px, p<2, p 3 are radii of three circles touching each other externally; and p is radius of circle through their centres ; prove that n (P 2 + Ps) P = 4V2.px.npx 3. If x, y, z are the distances of any point from the sides of a triangle, prove that 2 (x sin A) = 2 R n sin A 4. A circle circumscribes one triangle, and is inscribed in another : if the triangles are similar, prove that any side of inner : homologous side of outer = 4TI sin — : 1 5. If P, Q are any points in BC ; and p 1} p 2 , p 3 the respective circum-radii of ABP, APQ, AQC ; prove that R p % = p x p 3 . 6. In the figure of Euclid iv. 10, show that, approximately, area large circle : area small circle = 1 809 : 500 7. Prove that — (1) 2 (area BIC/area BE 2 C) = 1 (2) 2 (a/area BE X C) = 2/r Remarkable Lines and Circles 259 8. If A = 6o°, show that the bisector of A goes through N. 9. If AS meets BC in D, prove that DS = R cos A/cos (B - C) 10. If AS meets BC in D, and the circum-circle in D', prove that 2(DD'/AD) = 1 ' W.J. C. Sharpe: E. T. xl. 11. If Pi, p 2 , Ps are the radii of the in-circles of BSC, CSA, ASB respec- tively ; prove that 4R 2 2pj + 2RSap x = abc 12. Show that r lt r 2 , r 3 are the roots of x 3 — (4 R + r) x 2 + s 2 x — r s 2 = o 13. Prove that the area of the triangle formed by joining the points of con- tact of the in-circle is A r/(2 R). 14. Find the sides and angles of the principal ex-central triangle; and prove that its area may be expressed as 2 R A/r, or sy( 2 II cos — ) , or 8 s R 2 A/(abc), or A /A II sin — \ or 8R 2 IIcos — , or i abc 2 2 * {C'S-t} Show also that the radius of its in-circle is 4 R II cos — /E cos — 2 / 2 15. Prove that the area of the ex-central triangle IE 2 E 3 is 2Rs x . 16. Prove that 2 NA 2 = 3 R 2 - NO 2 . Note — Recollect that N, O, A, B, C are the circum-, in-, and ex-centres of the pedal A ; and that the circum-rad f of that A is R12 : then use results (5) and (18) of Example 4. 17. If x, y, z are the perpendiculars from A, B, C respectively, on the direction of SI ; prove that ax +by+CZ = o, where the one of x, y, z, which is on the opposite side of SI from that of the other two, is considered negative. Pet' CamV; '49. S 2 i6o Trigonometry — Chapter XIV 18. P is a point within a triangle ; prove that — if, i°, PA cos BPC = PB cos CPA = PC cos APB, then P is the in-centre ; and if, 2°, PA sec BPC = PB sec CPA = PC sec APB, then P is the orthocentre. 19. If P is a point within a triangle, at which BC subtends the angle 6, and if PS produced meets the circum-circle in Q, prove that SQ 2 - SP 2 = 2 (cot A - cot 6) A PBC Genese : E. T. lv. Deduce Chappie's Theorem that SI 2 = R 2 - 2 R r 20. In fig' (2) (page 233) of the 'ambiguous case,' if the angles ACB 2 , B 2 CN, NCBj are equal (OC being their common value) and if r, r 2 have their usual meanings for triangle AB X C, and K, r',, are corresponding quantities for AB 2 C, prove that = 2 COS a rr/+ K r 2 rrZ-K r 2 Note — Use forms (3) for r, r lf in Examples 2 and 3. 21. In the 'ambiguous case' prove that — (1) r r 7 : r^ r/ = b — a : b + a (2) I, I', E 3 , E 3 ' are concyclic ( 3 ) r + r / = K + r x = CN (fig' p. 233) (4) r K r 2 r 2 ' = A A' Tucker'. E. T. xxvii, xxx, xlviii. 22. If AN meets BC in OC, prove that— (1) AN : N OC = cos (B - C) + 2 cos A : cos (B - C) (2) B OC : OC C = sin 2 A + sin 2 B : sin 2 A + sin 2 C 23. Prove each of the following — (1) OS 2 = R 2 (i -8lIcosA) (2) Ol 2 = 4 R 2 (^811 sin 2 — -IIcosa) (3) OE^ = 4 R 2 (8 sin 2 — cos 2 -?- cos 2 — - IT cos AJ (4) IG 2 = |R 2 (i + IIcosA)--jRr + f r 2 Remarkable Lines and Circles 2,61 24. Prove that the distances between the in- and ex-centres of the principal ex-central triangle are respectively ^ ■ B + c o^- c + A nn . A+B 8Rsm— , 8Rsm , 8Rsin 4 4 4 25. From any point P perpendiculars PX, PY, PZ are dropped on the sides of a triangle ; prove that 2 area XYZ = (PS 2 * R 2 ) TT sin A. Interpret the geometrical meaning when P is on the circum- circle. When will XYZ be maximum ? 26. If p is the radius of the polar circle of a triangle, show that p 2 = — 4 R 2 cos A cos B cos C Account for the negative sign. 27. Show that the polar circle cuts the circum-circle and N. P. circle each at the same angle 0, where cos 2 6 = — cos A cos B cos C. 28. If r,, r 2 , r 3 are the roots of x 3 — px 2 + qx — t = o, show that the ex-radii of the pedal triangle are the roots of (pq - t) 3 x 3 - 2 (pq - t) 2 (q 2 - pt) x 2 + 16 qt 2 (pq - t) x - 8 1 2 {4 q 3 - (pq + t) 2 } =0 Edwardes : E. T. xl. Note — Use Example 5, and its Cor r ; and also results (11) (10) (7) (6) (4) of Example 4. 29. If SIO is a right angle, prove that 4 ( 2 R- r) (R + r) = 2a 2 30. If p is the in-radius of the pedal triangle, prove that — (1) Ol 2 = 2 (r 2 - Rp) (2) OE^ = 2 (r 1 2 - Rp) (3) OS 2 = R 2 - 4 Rp 31. Prove that the sum of the squares of the distances of the centroids of triangles BIC, CIA, AIB, BEjC, CE 2 A, AE 3 B from G is f 2ab + f R( 2 R-r) S. Watson : E. T. xxxiii. 262 Trigonometry— Chapter XIV 32. If r lf r 2 , r 3 are in H. P. ; and r, r 2 , R in G. P. ; show that A C i cos B = 4» that R = 9 r> and that tan — tan — = -• y 223 33. P is any point within a triangle, and p 1 , p 2 , p 3 the circtmi-radii of BPC, CPA, APB; show that (J*-)\(± + lL + °.)(_± + }L. + JL\ \Plp2pJ \Pl P 2 Ps/ V Pi P2 Ps/ VPl P2 Ps/ VPl P-2 Ps/ Note — Use the 2nd result of % 27. 34. Sj, S 2 , S 3 are the images of S with respect to BC, CA, AB respec- tively : show that if the circum- circles of Si S 2 S 3 and ABC cut at an angle 6, 2 cos 6 = 1 — 8 IT cos A 35. X, Y, Z are the centres of three circles touching each other externally in pairs, and with respective radii x, y, z : P is the centre of the fourth circle, touching and including the three : if (X, j3, y denote PX, PY, PZ respectively, prove that *© ,+ — s @ Math' Tri'\ '64. Note — Use Exercise 1 of p. 79. 36. If Al, Bl, CI cut SO inQ l5 Q 2 ,Q 3 respectively; and if cos A, cos B, COS C are in A. P. ; prove that SQ 1? SQ 2 , SQ 3 are in H. P. Pet' Camb'\ '66. 37. L, M, N are the points where the altitudes of A, B, C cut the sides of the pedal triangle : prove that OL cos (B - C) = OM cos (C - A) = ON cos (A - B) and that, if p is the in-radius of the pedal triangle, V P 2 _ T , „ TT P = I + 2ll OL 2 OL Pet' Camb': '66. 38. If p l5 p 2 are the radii of the circles touching b, c, and the circum- circle, internally and externally respectively, prove that A P2-P1 = 4Rtan 2 — Remarkable Lines and Circles 263 39. If x, y, z are the respective distances of I from E l5 E 2 , E 3 , prove that 32 R 3 — 2 (x 2 + y 2 + z 2 ) R — xyz = o 40. If x, y, z are the respective distances of O from A, B, C, prove that xyz (ax + by + cz) 3 + a be (x 2 + y 2 + z 2 ) (ax + by + cz) 2 — 4 a 3 b 3 c 3 = o 41. If I, m, n are the segments of the pedal line of P, included within the angles A, B, C respectively; and if Sa 2 = 2 o -2 ; prove that — (1) Va 2 - I 2 + Vb 2 - m 2 + Vc 2 - n 2 = o the negative sign being taken for one of the radicals (2) 2{(o- 2 -a 2 )l 2 } = 4 s 2 (3) 2 {(o- 2 - a 2 ) a 2 PA 2 } = a 2 b 2 c 2 Editor: E. T. xxix. 42. With a, b, c as diameters circles are drawn ; and b is the diameter of the circle touching, and including, all three : prove the relation Vb/s x - 1 + Vh/s 2 - 1 + */b/s 3 - 1 = Vs/^b-s) Math' Tri'\ '73. 43. If a triangle is acute-angled can OA, OB, OC be taken as sides of a new triangle? If this is possible ; and OC, j3, y are the angles of this new tri- angle, respectively opposite the lengths OA, OB, OC ; prove that / V COS 0C\ yj . (1 + 2; - 1 = n sec A \ cos A/ 2 1 1 COS A/ Math' 7ri': '81. 44. Pi is the in-radius of the triangle formed by joining the points of con- tact of the ex-circle E x ; and p 2 , Pz are similar quantities: prove that p 1 (1 - tan — j = p 2 (1 - tan -— ) = p 3 A - tan — -j Tucker : E. T. xviii. 45. If P\ is the in-radius of triangle E x BC, and p 2 , p 3 are the analogous radii, prove that v Pi P2 Pz v r x v r 2 v r 3 6*. Watson : E. T. xix. 264 Trigonometry — Chapter XIV 46. If P is any point on the circum-circle, prove that 2 (PA 2 sin 2 A) = 4 A and deduce that, if Q is any point on the N. P. circle, 2{QA 2 sinAcos(B-C)} = R 2 ( 4 II sin A + IT sin 2 A) Wolstenholme : 579. 47. Three circles are so placed within a triangle that each touches two sides and the two remaining circles : if x, y, z are the respective radii of the circles touching the sides of the angles A, B, C, prove that x ( 1 + tan — J = y (1 + tan — J =z[i+ tan — J = r ( 1 + tan - — J (1 + tan — J ( 1 + tan — j Note — This is the trigonometrical statement of what is known as Mal- fatti's Problem — a geometrical proof of which by Dr. Hart is given in Vol' I of the Q.J. Trigonometrical solutions are given in ' Gaskhi's solutions of the Geometrical Proble?7is proposed at St. John 's, Cambridge,' p. 82 ; and in Hymer's Trigonometry, p. 153. Putting 2 a, 2 /3, 2 y for A, B, C, the fig' gives y cot (3 + z cot y + 2 Vyz = r (cot (3 + cot y) and two sim'r eq'ns : fro?n these Vx cos OC + Vz sin a = Vy cos j3 + Vz sin (3 and two sim'r eq'ns : the rest follows easily. 48. Prove that 10. IS cos SIO = Rp_R r + r 2 where p is the in- radius of the pedal triangle. Also, by finding a similar result for 10 . IS sin SIO, show that 4 (area SIO) 2 = R 2 r 2 + 6 R 2 r p - R 2 p 2 - 2 R 3 p - 2 R r 3 - 2 R r 2 p - p 4 Editor : E. T. xxx. 49. Apply the result of Example 5 on p. 210 to show that the area of the triangle SIO is D2 .A-B . B-C . C-A — 2 R 2 sin sin sin 222 § 75. Some fundamental trigonometrical properties of the Le- moine and Brocard Geometry. Remarkable Lines and Circles 265 Examples 1. If through A, B, C lines AXY {or XAY), BYZ {or YBZ), CZX (or ZCX) are drawn, so as to ?nake the same angle {measured one way round) with AB, BC, CA, respectively, and forming a triangle XYZ ; then triangles XYZ, ABC are similar, and any side of one : homologous side of other = sin(co + 0):sinco where co is the Brocard angle of triangle ABC. A A A A A A It is at once obvious that X = A, Y= B, Z = C. Also YZ BZ + BY BC BC BZ BY BA BC ± BA BC _ sin (C + 6) sin sin C sin C ~ sin B sin A = cos ± sin . 2 cot A = cos ± sin cot co YZ : BC = sin (to + 6) : sin co Cor' (1)— When = co, and AX &c are drawn within ABC, the A XYZ shrinks up into a Brocard point. Cor' (2) — When = 7T/2, the ratio becomes cot co. Cor' (3) — When = co, and AX &c are drawn without ABC, the ratio becomes 2 cos co. Note — The particular case of this theorem, when the A XYZ shrinks into a point, probably gave rise to the discovery of the Brocard points ; and thence to the rest of this remarkable branch of modern Geometry. 266 Trigonometry— Chapter XIV 2. Magnitude of the Brocard Angle. A Agl \ If 12 is the pos' Brocard p't of the /j \. A ABC, then / /o \. AAA / J±L^_^ \ co = 12 AB - 12BC = 12 CA B C sin ( A — co) 12C 12 C 12 B _ sin co sin co sin co 12A 12B 12A sin (C — co) sin (B — co) .-. sin 3 co = sin (A — co) sin (B — co) sin (C — co) whence, as in Example 7, page 83, cot co = 2 cot A (1) or cosec 2 co = 2 cosec 2 A (2) other useful forms of this are 1 + IT cos A 001(0 = -,-,- — — (3) ITsinA K0J . v 2 R cos A R v AN 2a 2 cot co = 2 =^2(2bccosA) =4 ^ . . . (4) Othenvise — The following is an elegant proof* of this important property Let O which touches AB at A, and goes thro / C, cut the || to BC thro' A in P. Then PB cuts this O in 12, f A so that PBC = co. Drop AD, PN X s on BC. C N * Due to Mr. R. F. Davis, who kindly sent it to me immediately on its discovery in August, 1890. R. C. J. N. f See Euclid Revised, p. 394. Remarkable Lines and Circles 26y Then, if Q is the p't in which BC cuts O, A PC Q is a symmetrical trapezium .-. AD - PN, and QD - CN. .-. BN = QD + BD + CD Dividing each side by AD (or PN) and noticing that BAC = APC = AQB we get cot 03 = cot A + cot B + cot C 3. Length of the radius R x of the first Lemoine {or T. R., i. e. triplicate ration circle. A If KD, KE, || 8 to the sides AC, AB, meet AB, AC respectively in D, E ; and if U is the cross of KA, DE ; then D, E are p'ts on the T. R. O ; and T, the mid p't of KS, is its centre. Also, from the geometry of that A O, DTU = oj. B C .-. R x = TD = TU sec go = 2 sec co 4r. Length of the radius R 2 of the second Lemoine {or cosine) circle. A If an anti-|| thro' K to AB meets BC in D, then D is a p't on the O, K being its centre. Now (See § 73, Ex' 2, Cor') J_ from K on BC = 2 a a/2 a 2 R2 = KD = sinA 2 a* = R tan 00 (See § 75, Ex' 2) Cor'— 4 Ri 2 - R 2 2 = R2 (sec 2 co - tan 2 co) = R 2 268 Trigonometry— Chapter XIV 5. Length of the radius R 3 of the Taylor circle. If XYZ is the pedal A ; OL, /3, y the mid p'ts of its sides YZ, ZX, XY, respectively; and if CX/3 meets BC in D ; then D is a p't on the O ; and t, the in-centre of A OLQy, is its centre. Drop tn 1 on OC/3 ; and put = TT tan A Note — (p is the A subtended by any one of the 3 anti-]| s to the sides of ABC, formed by producing the sides of Oifiy, at any p't of the O lying on the same side of that anti-|| as t. If (f)' is the A subtended on the side opposite to t, we have tan (/>' = tan (it - .*. — sin (A — co) = sin co a .'. 1212' = 2 R sin co Vi — 4sin 2 co 8. Length of the radius p of the Brocard circle. This O has SK as diarr/, and goes K thro' 12 and 12'. A Also 12SX2' = 2 co. Remarkable Lines and Circles 271 aw 2 sin 2 a) R = — sec 0) Vi — 4 sin 2 00, by last Example = — Vsec 2 co — 4 tan' CO = — , Vi — 3 tan 2 co • 21 Cor*— p 2 = — sec 2 to - R 2 tan 2 co 4 = R x 2 - R 2 2 .'. 7T R x 2 = 77 R 2 2 + IT p 2 .*. area T. R. O = area cosine O + area Brocard O. Note — The centre of the Brocard O is T, the centre of the T. R. O. Exercises 1. If x, y, z are the perpendiculars from K on the sides of the medial triangle, prove that a 2 x /cos A = b 2 y /cosB = c 2 z/cos C Deduce Schlomilch's Theorem — If D is the mid' point of BC, and d the mid' point of the altitude of A ; and if E, e and F, f are similar pairs of points ; then D d, Ee, Ff cointersect in K. 2. On the sides of triangle ABC, triangles PAB, QBC, RCA are described external and similar to itself; so that the angles adjacent to A, B, C are respectively C, A, B ; prove that the circles round the three new triangles cointersect in the positive Brocard point of ABC. 3. If KP, KQ, KR are perpendiculars from K on BC, CA, AB respectively ; prove that — (1) PB/a + QC/b + RA/c = | - PC/a + QA/b + RB/c (2) PB . QC . RA + PC . QA . RB = (3 abc tan 2 aO/4. 272 Trigonometry — Chapter XIV 4. Prove that the tangent from A to the T. R. circle is R tan co Vb 2 + c 2 /a. 5. Prove that the perpendiculars from the centre of the T\ R. circle on the sides are proportional to cos (A — co), cos (B — co), cos (C — co). 6. If Pi, P2» P3 are the circum-radii of A 12 B, B 12 C, CX2A, show that p x p 2 p 3 = R 3 . Tucker. 7. Show that the Taylor circle intercepts on the side AB a length a cos A cos (B — C) 8. If L, M, N are the feet of the perpendiculars from 12 (or 12') on the sides of the triangle, show that area LMN = A sin 2 co. 9. If t is the centre of the Taylor circle, prove that — (1) the perpendicular from t on BC = — {cos A - cos 2 A cos (B - C)} (2)St:Kt=i+n cos A : - II cos A R. F. Davis. 10. If 6, (j>, X are tne angles made by KA, KB, KC with KS, prove that— V2 (b 2 + c 2 ) - a 2 (I) sin (B - C) . = Vi !(C 2 + a 2 ) - b 2 . sin (C-A) V2(a 2 + b 2 ) - c 2 sin(A-B) Sm * , x b 2 + c 2 -2a 2 x a c 2 + a 2 -2b 2 i , (2) ; — -=- — ^ r tan0 = - i -^ — 77^ r^-tand) VJ asin(B-C) bsm(C-A) r a 2 + b 2 -2c 2 1 = csin(A-B) tan ^ Simmons : E. T. 1. 11. OA, OB, OC are produced to OC, (3, y, so that AOC, B (3, Cy are respectively equal to BC, CA, AB : prove that — (1) triangles ABC, 0C(3y have the same centroid (2) area Otfiy = 2 A (2 + cot co) (3) a/3 2 + /3y 2 + y a 2 = 8 A (3 + 2 cot co) (4) the Brocard angle of (Xj3y = cot" 1 (2 cot co + 3)/(cot co + 2) Neuberg. CHAPTER XV Quadrilaterals and other Polygons § 76. The only new principle involved in this chapter is the idea of a negative area. Otherwise it will consist of Examples and Exercises on preceding results applied to Polygons. Def — If a point, in going continuously and completely round the boundary of a closed figure, crosses its own path anywhere, that path is called an autotomic (self-cutting) circuit : the sign of any area enclosed by an autotomic circuit is considered positive if that area lies on the left-hand of the travelling point, but negative if it lies on the right-hand; and the total area of the whole circuit is the algebraic sum of the several areas which it encloses. Illustrations (0 V Let fig' (i) represent the trace made by a skater, in going round a ' figure of eight ' on one foot, as indicated by the arrow heads : then the right-hand loop is a pos' area, and the left-hand loop a neg' area. Let fig' (2) represent a cross-quadrilateral, whose sides are taken in the order indicated by the arrow heads ; then the left hand A is a pos' area, and the right-hand A a neg' area. T 274 Trigonometry — Chapter XV Examples 1. Area of any quadrilateral (convex, or crossed?) in terms of its sides, and a pair of opposite angles. Let £2 be the area of a quad' ABCD ; d, e, f, g its successive sides AB, BC, CD, DA; x its diag' AC ; a its semi-perim' ; 6, .-. d 2 + e 2 — f 2 — g 2 = 2 (de cos 6 —fg cos (j)) Also 4 £2 = 2 (de sin Q + fg sin <£) Whence, by squaring and adding, we get 16 & 2 + (d 2 + e 2 - f 2 - g 2 ) 2 = 4 {(de) 2 + (fg) 2 - 2 defg cos (0 + (£)} = 4{(de + fg) 2 - 2 defg (1 + cos 6 + i.e. Vdefg- sin +4> ^ 2 , let the quad' be crossed, so that ABCD is an autotomic circuit — the sides AB, CD crossing in O. Then 12 = A AOD - A COB = A ADC- A ABC Now f 2 + g 2 — 2 fg cos <|) = x 2 = d 2 + e 2 — 2 de cos .-. f 2 + g 2 — d 2 — e 2 = 2 (fg cos $ — de cos 6) Also 4*0, = 2 (fgsin <£ — de sin 0) Whence, by squaring and adding, we get 16 Q? + (f 2 + g 2 - d 2 - e 2 ) 2 = 4 {(fg) 2 + (de) 2 - 2 defg cos (0 - )} = 4 {(fg - de) 2 + 2 defg (1 - cos $ - )} \ i612 2 = 4(fg - de) 2 - (f 2 + g 2 - d 2 - e 2 ) 2 + 16 defg sin' .-. 12 = 0- ~ r 2 + i 4. A quadrilateral is inscribed in a circle, radius R, and circumscribes a circle, radius r : if d is the distance between the centres, prove that the rect- angle under the diagonals is 8 R 2 r 2 /(R 2 - d 2 ) Prove also that, if 12 is the area of the quadrilateral, a its semi-perimeter, and OC, (3, y, b its angles, then cr/S cosec (X = r — £l/a 5. If R is the radius of a circle circumscribing a regular pentagon, each of whose sides is a, prove that R : a = 17 : 20 nearly 6. Show that the area of any polygon ABCD &c, circumscribing a circle is i\ its perimeter) 2 /2 cot — 7. Show that the area of a regular sixteen sided polygon, inscribed in a circle of radius unity, is V 2 — V 2 + V2 8. If a, b, c, d are the successive sides of a convex quadrilateral; and if a, c, produced, cross at an angle 6, and b, d, produced, cross at an angle c 2 sin c|) b 2 *«* d 2 27. A convex quadrilateral has two adjacent sides equal, and the other two equal (Professor Sylvester s Kite) and e is the ratio of two unequal sides ; if is the acute angle between a pair of adjacent unequal sides, and (p the angle between a pair of opposite unequal sides ; prove that . (f> i-e tan — = tan — 2 1 + e 2 Sylvester : E. T. xxiii. 28. If a, b, c, d are the successive sides of any quadrilateral; x, y its diagonals ; and (p the sum of a pair of opposite angles ; prove that x 2 y 2 = a 2 c 2 + b 2 d 2 — 2 abed cos (p R. U. Schol'\ '86. 29. A series of n circles, each of radius a, is ranged touching each other, so that their centres are concyclic : find the area enclosed within the external curvilinear polygon formed by their circumferences. 30. ABCDE is a regular pentagon : X, Y, P, Q are the respective mid points of AB, AE, BC, DE : I is the in-centre of the pentagon : if Al crosses XY, PQ in M, N, prove that 2 (IM — IN) = the in-radius. 31. The diagonals of a quadrilateral are equal, and at right angles : if the sides are a, b, c, d, show that its area is i {a 2 + c 2 + V 4 b 2 d 2 - (a 2 - c 2 ) 2 } and that this is expressible in terms of any three of the sides. What does the area become if the quadrilateral is cyclic ? Exercises 285 32. If a, b, C, d are sides of a quadrilateral circumscribing a circle; OC the angle between a, d ; and x the length of the tangent to the circle from the vertex of OC ; prove that (X (a + c) x 2 = ad (2 x — a + b) cos 2 — 33. P is any point within a quadrilateral ABCD : PA, PB, PC, PD are respectively denoted by f 1? f 2 , f 3 , f 4 : the respective perpendiculars from A, B, C, D, A, B on the bisectors of the angles subtended at P by AB, BC, CD, DA, BD, AC, are denoted by p l5 p 2 , p 3 , p 4 , p 5 , p 6 , prove that Pi Ps | p 2 p 4 = Ps Pe '1 '3 h "4 '1 f 2 and that, when P is at the cross of the diagonals, the dexter member becomes unity. Renshaw : E. T. xxv. 34. ABCDE is a cyclic pentagon : prove that EA 2 . BC . CD . DB + EC 2 . AB . BD . DA = EB 2 . AC . CD . DA + ED 2 . AB . BC . CA. Brill: E. T. xliii. Note — Deduce from Salmon's Theorem {Conies, 6th ed', p. 87) that, if 'A, B, C, D are concyclic, and E any other point, then EA 2 . A BCD + EC 2 . A ABD = EB 2 . A ACD + ED 2 . A ABC 35. With the notation of Exercise 28 ; show that, if x is conterminous with a, d, then jx 2 (ab + cd) - (ac + bd) (be + ad)} 2 sin 2 i- ' J 2 + {x 2 (ab - cd) - (ac - bd) (bc-ad)} 2 cos 2 $ = 4 a 2 b 2 c 2 d 2 sin 2 cj) Wolstenholnw. 597. 36. Show how the construction of a regular heptagon may be made to depend on the trisection of the angle cos -1 ( -) ■ V2V7/ Cay ley : E. T. xl. Note — If x = 4 cos 2 6, then sin 7 6, expressed in terms of x, will give x 3 — 5 x 2 + 6 x — 1 = o, where 6 is any A a side of the heptagon subtends at the in- centre. 286 Trigonometry— Chapter XV 37. If c 1} C 2 , c 3 , C 4 are the cosines of the angles of a convex quadri- lateral, prove that Sex 4 + 4Sc 2 2 c 3 2 c 4 2 + 8nc x = 22c 1 2 c 2 2 + 42c! 2 . II c x fe/ } Chrisfs and Emm', Camb': '90. 38. A quadrilateral ABCD is inscribed in a circle centre S, and circum- scribes a circle centre I : if SI produced meets the circum-circle in X, Y, prove that 11 1 1 LA 2 + ic 2 ~ Tx 2 + Iy 2 Trin' Camb'\ 'go. Note — Use result of Example 4. 39. Prove that the distance between the mid points of the interior diagonals of a cyclic quadrilateral, whose sides are d, e, f, g, is \ / {(d 2 -f 2 ) 2 eg + (e 2 -g 2 )df}/{4(ef + dg) (de + fg)} Sid' Camb': '91. 40. A cyclic pentagon ABCDE (sides a, b, c, d, e) circumscribes a circle ; prove that a (bri^i) + b fefinrb) + e (dTTrb) t d(-^)i,( b ~° )-, \e + c — d/ \a + d — e/ A. Russell : .£. 71 1. Note — tan — = area/a (cr — b — d) tan — = area/a (or — e — b) whence sinA — sinD = d — eye + a — b . sin (A + D) and sim'r results : then 2 R = —a/sin (E + C) = &c. 41 . A cyclic hexagon circumscribes a circle ; if R is the circum-radius, r the in-radius, and d the distance between the centres ; prove that r r sin cos -1 = j + cos sin -1 = -, = 1 R - d R + d Genese : j5\ 7". xxvii. Note — Take a position of symmetry, as in Example 4, and use Exercise 34, on p. 218, as a Lemma. The elementary solution given in the E. T. Reprint is wrong. CHAPTER XVI Application of Trigonometry to Surveying § 77. One of the most important applications of Trigonometry is to the practical work of Land-surveying. Before giving examples of such application, a few technicalities are here premised. Def — Any measured straight line is, in surveying, called a base. Def — If one point is visible from another, through a tube of small bore, then the angle round which that tube must be turned, in a vertical plane, to become horizontal, is called the elevation, or depression, of the first point with regard to the second — the former term being used when the first point is above the second, the latter when the first point is below the second. Def — The angle which the join of two points subtends at a third is the angle between the joins of the two points to the third. Def — The angle which any direction must be turned through to make it coincide with any Compass point is termed the bearing of that direction from that Compass point. Or, again, the word bearing is used to indicate simply the direction in which an object is seen, or in which it is inclined. Note — Thus it might be said that — (i) the bearing of a tower from the North is 15 ; meaning that the direction in which the tower is seen is inclined to the N-point of the Compass at 15 : (2) the bearing of a ship is N. E. ; meaning that the direction in which the ship is seen is the N. E. point of the Compass. (3) a wall bears 35 East of South ; meaning that the line of the wall makes an angle of 35 , on the East side of the S. point of the Compass, with the direction of that point. 288 Trigonometry — Chapter XVI In the Problems of Surveying (often rather absurdly called Heights and Distances) we are supposed to be able to measure bases, that is distances between accessible points ; and to observe (by instruments giving their magnitude) either the angular elevation of a point above, or its angular depression below, the horizontal plane through the eye ; or, again, to measure the angle which the join of two visible points subtends at the eye — such measurement and observation being made by means of the practical instruments used by Surveyors. These instruments are Gunters Chain, Newton's* Sextant, the Theodolite, the Vernier, the Spirit Level, and some others of less importance : descriptions of them will be found in any Treatise on Surveying. The Student will be here supposed to know the Points of the Mariner s Compass : as there are 3 2 such points, the angle between any two consecutive points is 360° 732, that is n°f. These things assumed, the Problems of Land- Surveying are merely applications of the trigonometrical solution of triangles. Examples 1 . To find the height h of an inaccessible point — say the summit of a mountain — above the horizontal plane in which the obseiijer is situated. Let S be the summit, P the position of the observer. Let the observer make the following measurements. i°, a base (a say) between P and any well defined p't Q * Not Hartley's — as it is generally termed— see HerscheVs Astronomy, io.th ed / p. 116. Trigonometrical Surveying 289 A 2°, the elevation OC of S at P : A 3°, „ (3 which SP subtends at Q : A 4°, „ y „ SQ „ „ P. T , . OD . _. a sin a s'm (3 Then h = SP sin OC = — = — -= i- sin (/3 + 7) Note—\i Q can be taken in the join of P to the foot of h, only two A s need be observed. 2. To find the distance apart of two inaccessible points — say the summits of two mountains. Si Let S x , S 2 be the summits. Let the observer measure — i°, a base (a say) between two well-defined p'ts A r B ; not necessarily in the same plane with S x , S 2 : A 2 , the OC which Sj A subtends at B : 3°, >) A >) SiB >j »> A: 4°, )) A 7 3> SjA » » B: 5°, }} A S }) S 2 B » }} A. Then s, L A = a sin OC sin (OC + /3) s 2 A = a sin y o!« f«, _L *\ whence, as the A which S x S 2 subtends at A can also be observed, the solution of the A Sj AS 2 is brought under the cond'ns of the case solved on p. 230. U 290 Trigonometry — Chapter XVI 3. A, B, C are three Stations, whose distances from each other are known : to find how an Observer, at a fourth Station P, can calculate his distance from each of the three* {Hipparchus' Problem} A A BPC = OC, and APC = /3, can be observed. A A Let CAP = 6, and CBP = (p. Then, denoting the sides and /\ s of A ABC as usual, we have 6 + tan .'. 6 — (j) is known Q and c/> are known * Sometimes called by Surveyors the Stasimetric Problem — especially useful in Maritime Surveying : it was originally solved by the ancient astronomer Hipparchus : see Philosophical Transactions, vol' vi, p. 2093. The Problem, and some cognate Problems, are very ingeniously treated in a volume by the late Professor Wallace of Edinburgh. Trigonometrical Surveying 291 _ A . sin (0 + /3) . ■ \ PA = b ^ — t^^j is known sin fi PB =a and PC = a sin ($ + a) sin OC sin cj) sin a' > jj >> _/\^^ — When P is concyclic with A, B, C, we have 6 = 5 miles at most, and is usually much less, the error by assuming that x = V2 R h, and sin 9 = d will not be great : then (R + h) 6 = V2 R h and .-. 8 - R V2 R h/R + h For small heights, R/(R + h) = 1, nearly and then 8 = V2 R h, the formula generally used. U 2 292 Trigonometry — Chapter XVI Cor* — If we assume that R = 3960 miles, which is very near the truth, then .'. b 2 = 2 h x 3960 .". 2§ 2 = 3 h x (1760 x 3) whence we get the following fairly approximate rule — Distance of horizon in miles = ■y ■% x height of object in feet. For example : the top of an object 150 feet high can be seen about 15 miles off. Def — The angle which a tangent to the Earth's surface, from a point above that surface, must be turned through (in a vertical plane) to become horizontal, is called the dip of the horizon at that point : it is evidently equal to the angle which the tangent subtends at the Earth's centre. 5. A right pyramid, on a square base, stands on a horizontal plane ; and, when the sun has an elevation (X, the distajice of the extreme point of its shadow from the corners of the base nearest to it are a, b, c : show that the height of the pyra??iid is a tan OC sin cf) cosec 6 where 6 and (f) are given by the equations h 2 — a 2 tan (45 - 0) = tan 5 (♦ * 9 c 2 -a 2 tan - tan \— + tan -1 s— 2 / 2 2 a 2 ' Sun; Pef CamV\ 50' Let V be the vertex of the pyramid ; S the extremity of its shadow. Let A, B, C be the corners of the base whose respective disf's from S are a, b, c. Then, if VN is the alt' of V, N is the cross of the diag's of the base. Trigonometrical Surveying 293 AAA Now if VSN = OC, SAN = , SNA = 0, then VN = SN tan OC = a tan a sin $ cosec Also BC = 2 AN = 2 a sin ((f) + 0) sin A _ ,-sin (rf> + 0) AB = a V2 ^- = AC sin b 2 = a 2 + AB 2 - 2 a AB cos (<£ - 45 ) .0 - „ sin 2 (A + 0) sin ( + 0) "•-■•'""-OT--' 8 ' Ig j (cos 4, + sin,/,) b 2 - a 2 _ sin (<£ + 0) /sin (f> cos - sin Oxf 1st Public Exam': '74 zg6 Trigonometry—Chapter XVI 16. A rock is observed from the deck of a ship to bear N. N. W. ; and, after the ship has sailed 10 miles E. N. E., the rock bears due West ; find its distance from the ship at each observation. Chris? s Caml/: '46 17. While sailing S. W., a man observed two ships at anchor, one N. N. W., the other W. N.W., and, after sailing 5 miles, he saw the ships N. by W., and N. W. respectively ; find their bearing, and distance from each other. Christ's Camb f \ '45 18. AB (length a) and CD (length c) are two vertical posts at the edge of a river : P is the position of an observer on the edge exactly opposite A : if AB == AC, and the elevations of B and D at P are equal, find the breadth of the river ; and prove that cos APD = a 2 /c 2 = cos CPB 19. A balloon being supposed to ascend from the Earth with a given uniform velocity, at a given inclination to the horizon, and towards a given point of the compass, two angles of elevation are observed from the same place, at a given interval of time : find an equation to give the height of the balloon at the time of either observation, its angular bearing at that time being given. Joh' Camb'\ '33 20. The summit S of a mountain is just seen from a boat in a position A at sea ; the boat is rowed directly towards S, and when at B the elevation of S is OC : if the Earth is supposed a sphere centre O, and if SO cuts its surface in C, show that if arc AC = a, and arc BC = b, then approximately, a 2 — b 2 Earth's radius = ; — cot OC 2 b a 2 b and height of mountain == — ^ :-= tan OC s a 2 - b 2 Joh' Cam!/'. '33 21. Erom the summit of two rocks A, B, at sea, at a given distance 8 apart, the dips OC, ,/3, of the horizon are observed ; and it is remarked that the summit of B is in a horizontal line through the summit of A : show that the radius of the Earth is b/cos- 1 (sec OC cos /3) ' Pef Caml/'. '49 22. A square tower stands on a horizontal plane : from a point of the plane three of its corners have elevations 45 , 6o°, 45 , respectively : find the ratio of the height to the breadth of the tower. Ox* 1st Public Exam': 89 Trigonometrical Surveying 297 23. A flagstaff, on a tower, is observed from two points in the same horizontal plane, one due South, and the other due East of it, and subtends the same angle at each point : the elevations of the top of the staff at the same places are respectively tan -1 a, tan- 1 ^3: if c is the distance between the points of observation, prove that the height of the flagstaff is C (a/3 - i)/V0L 2 + /3 2 Ox/ ist Public Exam! ': '78 24. Two vertical walls are h x and h 2 feet high, and are mutually perpendi- cular : when the Sun is due South their shadows are observed to be a x and a 2 feet broad, respectively : if then 6 is the Sun's elevation, and ultimately Maxima and Minima 301 The treatment of inequalities is very similar to that of maxima and minima. The Student should keep in mind what is given in * Algebra under this head : especially he should recollect that a x + a 2 + a 3 + &c + a n > n Va x a 2 a 3 &c a n and that the Limits of value of the fraction ax 2 + 2 bx + c / lx 2 + 2 mx + n are given by the eq'n in y (b - my) 2 = (a - ly) (c - ny) Some examples of trigonometrical inequalities have already been incidentally given in §§ 44-49. Examples 1. If the distances of a point P, zvithin a triangle ABC, from A, B, C are x, y, z ; and the angles BPC, CPA, APB are 0, (p, X. — respectively in each case— find the maximum value of x sin + y sin (j) + z sin x and the position of P when this is the case. We have c 2 = x 2 + y 2 — 2 xy cos v = x 2 + y 2 — 2 xy cos (0 + fy) = (x sin 6 + y sin > » „ a> „ y cos (j) = z cos ^ and zsin^ + xsin0 » » 5 > b, „ z cos )( = x cos .•. the max/ value of 25 (xsin 6) is the semi-perim' of A ABC; and the position of P for which this is the case is when x cos = y cos cj) = z cos ^ i. e. when P is the in-centre of ABC. (See Ex r 18, p. 260) * See C. Smith's Algebra, Chapter XXVI ; or ChrystaVs Algebra, Chapter XXIV : in the latter the connection between inequalities and maxima and minima is brought out. 3 02 Trigonometry — Chapter XVII 2 . Given the angles and perimeter of a convex quadrilateral ; to find when its area is maximum. Catalan : E. T. 1. {Prof Genese's Solution) Let ABCD be the max* position of a quad' whose l\ s and perim' are given ; and let AB'C'D' be a con- secutive position. Draw Cm, Dn || to AB, BC respectively. The rest of the fig' is self- explanatory. By Fermat's principle, area ABCD = area AB'C'D' .-. area CDD'X = area BB'C'X' .-. (CD + XD') CX sin X = (C'B' + XB) C'X sin X Now ultimately CD = XD', and CB = C'B' = XB .-. XC/XC = CB/CD, ultimately Also, since the perim' is the same in all positions, BB' + C'X + D'n = CX + Xm + DD' XC^-5 + XC' +XC^ = XC + XC'^ +XC sin C sin B sin D sin B ' " sin D sin B + sin C - sin P _ XC sin B _ CB sin B sinC + sin D -sinQ ~~ XC'sinD ~ CD sin D CB.CPsinC CD.CQsinC ""■ PC + PB - BC ~ QD + QC - CD .*. O touching BC and PB, BC produced = „ „ CD „ QC, QD .'. these O s are the same i.e. quad' ABCD circumscribes a O \ its area is max' when AB + CD = BC + DA Maxima and Minima 303 3. If the sum of any number of varying angles is given, then the product [and the sum) of the sines of those angles will be maximum, when the angles are all equal. For if x + y = OC, a fixed A 2 sin x sin y = cos (x — y) — cos OC and .*. is max f when x = y Now let there be n A s > 1} 0%, 3 , &C n , whose sum is fixed : then, by the foregoing, if any two of them (say 1 , 2 ) are unequal, we can, without altering the rest or the total sum, increase the product sin X sin 2 , by making X , 2 equal. .'. IT sin is max? when X = 2 = &c = n SimTy 2 sin „ „ 5> 55 Note — Obviously II cos and 2 cos can be treated in exactly the same way. 4. If A, B, C are angles of a triangle, the minimum value of 2- cot 2 A is unity. This appears at once by considering that 2 cot 2 A = i + \ 2 (cot A - cot B) 2 and .*. is min f when cot A = cot B = cot C Cor' — The min' value of 2 tan 2 — is unity 2 5. If A, B, C are angles of a triangle ; and x, y, z any variables ; then 2 x 2 2 2 (yz cos A) or 2x 2 = 2 2 (yz cos A) 304 Trigonometry — Chapter XVII Exercises 1 . Given the sum of any number of angles, show that the sum (and product) of their tangents is minimum when the angles are all equal. 2. If a tan 6 + b tan (f) = c, where a, b, c are fixed positive quantities; and 6, d> are each less than a right angle ; prove that a sec + b sec (f) is minimum when 6 = tt/2. 5. Find what value of x gives the maximum value of — (1) sin 2 x sin (a — 2x) (2) sin — sin (3) m n sin x cosx 1 + sinx + cosx 6. In any triangle, except the equilateral, prove that 2 (a cos A) < s 7. Show that, if A + B + C = 180 , then 2sinA<^,but >6 V3~ITsin A 8. Show that 2 sin A ~* 2 COS A >, or < 1, according as the triangle is acute or obtuse angled. 9. If p is the radius of the N. P. circle, and r, s as usual, show that s 2 lies between 27 p 2 and 27 r 2 10. Prove that the sum of the cosines of the halves of any two angles of a triangle is greater than the cosine of half the third angle. Exercises 305 11. If the equation tan - = (a - i + tan 6) /{pi + i + tan 6) has real roots, show that CX 2 > I. 12. Prove that, for real values of x, the Limits of x 2 — 2 x cos OC + i /x 2 - 2 x cos /3 + i are I — COS OC / 1 — COS /3 and I + COS CX / 1 + COS /3 13. Prove that, in an acute-angled scalene triangle — (i) 2 tan A (or II tan A) > 3 V3 (2) 2 cot A > V3 (3) 2 (tan A tan B) > 9 (4) II sec A > 8 14. Show that the maximum value of the Brocard angle is 30 . 15. A circle, radius r, is inscribed in a sector, radius x, and chord y: if x + y remains constant, find when r is maximum and when minimum. 16. If a, b, c are the sides of a triangle in descending order of magnitude ; prove that, of the six squares which can be described so as to have their corners on the sides (or sides produced) of the triangle, area greatest square /c + b sin A\ 2 least uare _/c + b sin AV „ \c + bsin C/ 17. If D is the diameter of the circum-circle of a triangle ABC ; and Oi , §2? §3 t ne diameters of the greatest circles touching a side externally, and the circum-circle ; prove that VDg 1 8 2 6 3 = (abc)/(4s) 18. AB, BE, CF are the bisectors of the angles of a triangle ABC, termi- nated by the opposite sides : prove that — v ; A ABC * 4 64 A 3 (2) AD.BE.CF> (a + b) (b + c) (c + a) 19. CP, CQ are drawn to meet AB in P, Q, and to be equally inclined to the bisector of the angle BCA ; find when the circum-circle of PCQ is minimum, X 306 Trigonometry — Chapter XVII 20. The diagonals of a quadrilateral are a, b, and they form an angle Of ; prove that the area of the maximum rectangle circumscribing the quadrilateral is ^ab (1 + sin Of) 21. AB is the diameter of a fixed semi-circle, centre C ; P is a fixed point on the radius perpendicular to AB ; XY is a varying chord through P : prove that the area AXYB is maximum when the angles XCY, PAC are equal. Q. C. B. '73 22. If the altitude AN of a triangle is fixed, and its in-circle goes through its ortho-centre ; prove that its maximum in-radius is ^ AN. Knowles : E. T. xli Note — Prove I0 2 = 2 r 2 — AO .ON, and deduce. 23. A, Bare fixed points, and Pa point moving so that PA + PB remains constant : if 2 denote the angle APB, and if C is the point of AB dividing it into parts m, n ; prove that when CP is minimum its value is 2 mn cos / V(n\ + n) 2 sin 2 + (m — n) 2 cos 2 24:. If Of, (3, y are the lengths of the minimum lines bisecting a triangle, prove that V2(a 2 /3 2 ) = 2 A Murphy: E. T. xxv 25. XY is a chord of a fixed semi-circle ; XP, YQ are perpendiculars on its diameter : if XY varies in length, but is always parallel to a fixed direction, prove that the area PXYQ is maximum when XY subtends a right angle at the centre. 26. If x = b +csin0/a+ C COS0, where a, b, C are all positive, and C least, show that there is one value of for which x is maximum, and one value for which it is minimum ; and that both are given by a cos + b sin = — c. Q. C. B. '89 27. ABCDEF is a hexagon in which each of the sides AB, DE is 2 c ; and each of the sides BC, CD, EF, FA is 2 a ; prove that when the hexagon has its maximum area this area is 8 1 2(p+ Cf (p- C)^ where p is the positive root of the equation ' p 2 — cp = 2 a 2 Math' Tri': '87 CHAPTER XVIII Expansions — Series — Factors § 79- A digression is here introduced, which would have had its natural place in the Preliminary Chapter, had the necessity for it been foreseen in time. Algebraic* Note On the Convergence of Infinite Expressions Def — An expression, the number of whose terms is infinite, is said to be convergent, if its value, starting from any given term, when only a finite number n of its terms is included, converges, that is tends continually to a definite limit (not zero) as n increases indefinitely : otherwise such an expression is said to be non- convergent. In the term non- convergent are included the cases when, as more and more of its terms are taken, the value of an infinite expression — (i) diverges, that is increases without limit : (2) oscillates, that is has sometimes one value, and sometimes another : (3) tends to zero as its limit. In such expressions the infinite number of terms may be arranged as — (A) a sum, called a series ; or (B) a product : i. e. they may take either of the forms Uj + u 2 + u 3 + . . . or (1 + Ui) (1 + u 2 ) (1 + u 3 ) where u 1? u 2 , u 3 , &c may be pos' or neg'. * This note is intended to bring out some points which cannot be conveniently disposed of by a reference to text-books on Algebra. The only English text-book on that subject, wherein the question of Convergence will be found thoroughly discussed, from the point of view here put forward, is ChrystaVs Algebra ; but the discussion of it given there (in Chapter XXVI) is rather elaborate and difficult for the ordinary Student at this stage of his career : hereafter he should not fail to study it carefully. X 2 308 Trigonometry— Chapter XVIII We proceed to consider these forms separately. (A) Series. Def — If there is a limit to which a series tends continually, and from which it may be made to differ by less than any assignable quantity, by taking a sufficient number of its terms, this limit is called its sum to infinity ; and the quantity by which it differs from this limit, when only a finite number r of its terms is taken, is called its residue after r terms. Clearly a series is convergent if its residue may be ultimately neglected. For example, the geometrical series i + x + x 2 +x 3 + ...+x r_1 + . . . , where x < i, has for the sum of its 1st r terms i x i x , which — I— X I— X I— X and .". differs from 1/(1 — x) by the residue x r /(i — x). But, as r increases indefinitely, this residue diminishes indefinitely, and /. may ultimately be neglected. .•., as r increases, the sum tends more and more to become equal to 1/(1 — x), and finally differs from it by less than any assignable magnitude. .*. the series is convergent, and its sum to infinity is 1/(1 — x) It will J. be a sufficient test of the convergence of a series if we can show that, term for term, it < a G. P. whose common ratio < I. Again the series i+^+^ + x + «-- ma y De arranged in groups thus — (* + 1) + (* * i) + (i t i + I + i) + (4 -. • + tV) + ••• . Now each group > -| .*. the series > 1 taken as often as we please i. e. > a quantity which can be made as great as we please and .*. it is non-convergent. Taking a series of the form u x + u 2 + u 3 + . . . + u f + . . . where all the terms are pos', we see that obviously — i °, the terms must constantly diminish if it is to converge ; for otherwise its sum is constantly increasing : 2°, if it is conv't when all its terms are pos 7 , a fortiori it is conv't when some are made neg' ; for this change will diminish the numerical value of the residue. Note on Convergence 309 By comparing the series 2 u q with the G. P. 2r n , it is an immediate de- duction* that — if the Limit of u r + j/u r < I. when r = oo , then 2 u is convergent. Hence the fraction u f+ j/u is suitably called the test-ratio of convergence. In the preceding series the terms are all pos' ; but if the type of a series is w x - u 2 + u 3 + u f . . . . where the terms are alternately pos' and neg', the residue after r terms is and .\ lies between + u f + 1 and + (u f + x — u f + 2 ) provided the numerical value of each term < that of the preceding term : .*. if, after a certain term, each term < the preceding term, the series converges to a definite limit, provided that u f diminishes indefinitely as r increases. The above arrangement shows the closeness with which the limit is ap- proached when r terms are taken. For example, r terms of the series *-i + i-i +....± 1 -.... differs from log 2 by less than i/(r +2), and .'. it is convergent. Eut this is only true when the terms are arranged in order as above ; for if the series is arranged in form (i + S + * + + + •.•■)-& + I * i +.-.0 it represents the difference of two non-conv / t series, and .*. is itself non-conv't. Hence there are two types of convergent series — (a) those which are conv't when all the terms are made pos': (/3) those whose convergence depends on the accidental arrangement of the pos' and neg' terms. The former are said to be absolutely convergent : the latter (sometimes called semi-convergent) may appropriately be said to be tactically conver- gent. * See C. Smith's Algebra, p. 317 ; or ChrystaVs Algebra, vol' II, p. 107. 310 Trigonometry— Chapter XVIII (B) Products. if (i + u x ) (i + u 2 ) . . . (i + u r ) = n r and (i - u x ) (i - u 2 ) ... (i - u r ) = II' r where u lt u 2 , . . . u are all pos', we see at once that, if FL , II' tend to finite limits — i°, each of the quantities u l3 u 2 , . . ., u f < i ; and 2°, when this is the case, II may tend to a finite limit, or to oo , and II'j. „ „ „ „ , or to zero. To find conditions, change the forms thus : log IT r = log (i + Uj) + log (i + u 2 ) . . . + log (i + u r ) = "i - \ «V + i "i 3 ~ • • • + u 2 - \ u 2 2 + J u 2 3 - . . . .-. log rJ r lies between o and u x + u 2 . . . + u f .■. , provided Uj + u 2 + . . . + U is com/t, log IT is pos' and "jf* a definite quantity and then also IT r „ „ „ „ ,, i. e. FT tends to some limit less than a definite quantity, as r increases. .*. IT then converges. But if u x + u 2 . . . is non-con-/ 1, so is IT ; for (i + Uj) (i + u 2 ) . . . = i + u 1 + u 2 . . . + products of u 1} u 2 &c. Again - log IP = Uj + | u^ + | u/ . . . r 2 -r 2^ U 2 + -3 U 2 and .-. > u x + u 2 •»-... ". if Uj + U 2 . . . > a finite magnitude log (TI'j.)- 1 > finite mag 7 and .'. II' = o r Note on Convergence 311 So that then (in accordance with def) IP is non-conv / t. But if u 1 + U 2 . . . is conv't, then since logll' < — (1^ + u 2 . . .) .-. IT^ is conv't. Hence the convergence of IT and IT' is absolute or tactical as is that of Uj + u a + : . . Cor r — If either of II II' converges, so do the other. As Examples (1 + i) (1 + -|) (1 + -J) . . . is non-conv't (=00) for then IT r = (r+ 2) I2 C 1 - i) C 1 - i) (1 - t) • • • is non-conv't (= o) „ IT r = i/(r + a) Again (i+^(i+^)(i + *)... and (*-?)(*-£)(*-$)■- are each absolutely conv't, for ill 1 2 2 2 3 2 < 1 + V 2 + 2V + (? + ? + f 2 + ?) &C in Sim/r gr ° UpS 248 < ■ + ? + ? + p + • • • 1, each of the infinite expressions u x cos r + U 2 COS S (/) + ... and (1 + UjSin 1 ^)^ + u 2 sin (j)) ... is absolutely conv't, provided that u x + u 2 + u 3 ■+- . . . is absolutely conv't. 3^2 Trigonometry — Chapter XVIII § 80. Theorem — If n is a positive integer, expressiofis can be found for sin n X and cos n X in terms of sin X and COS x, where X is measured in any unit ; and hence can be deduced series for the sine and cosine of any angle in terms of the radian measure of that angle. Generalizing the law indicated in the expansions of COS (Oi + /3 + y) and sin (06 + /3 + y) on pp. 66, 67, assume that, if k = OL x + 0i 2 + ... + 0L a , cos k — 2 — 2 2 + 2 4 — ... sink = 2,-^ + 2 5 -... where 2 r means the sum of all the products for??ied by multiplying together the sines of any r of the angles and the cosines of the remaining (n — r) angles. Introducing a new angle A, we have cos (k + X) = (2 -2 2 + ...) cos X-(2 1 -2 3 + ...) sin X = 2 cos A — (2 2 C0SX + 2J sin X) + ... where 2/ is the same for (n + 1) A s that 2 r is for n A a - sin (k + X) = (2 1 — 2 3 +...) C osX + (2 — 2 2 + ...) sinX = (2j cosX + 2 sinX)— (2 3 cosX + 2 2 sin X)+ ... = 2'— 2 ' + 2'— ... .-. if the assumptions are true for any particular value of n they are also true for the next greater value. But they are true when n = 3. .*. , by the Law of Induction, they are true generally. From these, if 0i x = 0L. 2 = ... = 0C n = x, we get that cos n x = c n - n C 2 c 11 - 2 s 2 + n C 4 c n ~ 4 s 4 and sin nx = n 1 c n ~ ] s — n C 3 c n " s 3 + n C 6 c n ~ r ' s n - 5 s 5 - ... J Expansions 3*3 where sin x = s, cos x = c and n C r = the N° of comb'ns of n things r together. Now suppose that nx = a radians, and that this equality is maintained, when n increases, by a corresponding diminution of x, so that OL remains unchanged. Then OL cm hi COS OL = COS n n (■-m +(-*)(-!) (-*)?; COS n— 2 COS 1 06 n a Consider a very great number r of terms, and let n be chosen to be an integer itself very great compared with r. Then, as sin 6 / < i, and cos 6 < i, the numerical value of each term of this series < „ ,, ,, the corresponding term of the series a 2 06 4 1 r+ — r- 0L G + 2 ! 4 ! 6 ! But this last series is absolutely convergent ; since, when all its terms are made pos', its test-ratio (r + i)th term _ 06 2 rth term 2 r (2 r — 1) and the Limit of this (when r = 00 ) < 1, for all finite values of OL. .'. we may ultimately neglect all its terms after the rth. .*. , as the convergence is absolute, we might do the same with the 1 st series, even if all its terms were pos', and a fortiori as it stands. .*. cos OL approximates to the sum of the 1st r terms of the 1st series. 3H Trigonometry— Chapter XVIII Now sin OL n OL and cos n_r — , and also the coeff , each ulti- n 06 L n J mately become unity, as n increases towards oo . OL 2 Ot 4 a 6 .-. cos a = i + — - — — + ... 2 ! 4 ! 6 ! In a precisely sim'r manner we should find that sin OL oC ol 5 a 7 OL T + -T i + • 3! 5! 7! {Newton) Cor f (i> . 77 X I /fX\ 3 sin X° = — : I -5- ) + • • . 180 3 ! \i8o/ cos X 2 ! V180/ 4'! \i8o/ C^ (2) — By actual division, we have (X 3 2 a 5 tan OL = OL + — + + • . - 3 i5 Applying the principle of Reversion of Series to this (see ChrystaVs Algebra, vol' ii, pp. 254 . . .) we get, as successive approximations OL = tan OL OL = tan OL - i tan 3 OL OL = tan a - \ tan 3 (X + \ tan 5 a From the 3rd of these, by putting x for tan OL, we have the approximation y3 v5 tan -1 x = x h , when x is small 3 5 This suggests that tan -1 X may be expanded in powers of x, if it is between limits which make the series conv't. § 81. Def — If (x) = (— x) then (x) is called an even function of x ; and if (x) = — (— x) then (x) is called an odd function of x. Expansions 3 L 5 Lemma — If a function of X can be expanded in a series of ascending powers of x, then this series will contain — i°, for an even function only even powers of x ; and 2°, „ odd „ „ odd „ „ ; For if (f)(x) = 0i + OLi X + 06 2 X 2 + 0L 3 X 3 + ••• then (f)(— x) = 06 — OL x X + 06 2 X 2 — 06 3 X 3 + ••• .-. i°, if 0(x) = 0(- x) we have, by subtraction oq x + a 3 x 3 + 06 5 x 5 + ... = o whence 0^ = 0, 0i 3 = o, 0i 5 = o, ... and, 2 , if 0(x) = — rf>(— x) we have, by addition a + a 2 x 2 + a 4 x 4 + ... = o whence 0i = o, 0i 2 = o, 06 4 = o, ... Note — Of the six T. F.s, obviously cosines and secants are even, but sines, tangents, cotangents, and cosecants are odd functions of angle ; so that the expansions we have found for sin OC and cos OC are in accordance with the preceding theorem. Now, although it is entirely illegitimate to assume that a function can be expanded in an infinite series of whose convergence we know nothing, yet the provisional assumption of such an expansion in a particular case, may be a guide towards finding a rigorous proof of its existence. In accordance with this idea let us assume provisionally that tan -1 x being an odd function of x, can be expanded as a con- verging series in the form a x X + Ot 3 X 3 + 0L 5 X 5 + ... x -(- y then, since tan -1 = tan -1 x + tan* 1 y, *we have 1 - xy * For this method I am indebted to Prof Genese. 3 X 6 Trigonometry— Chapter XVIII x + y /x + y\ 3 Oj + 06 3 ( i- ) + ... 1 — xy \ 1 — xy / = 06, (x + y) + oc 3 (x 3 + y 3 ) + Equating coeff's of x 2r y, i.e. of x 2r_1 (xy), we get (2 r - 1) a 2T _ x + (2 r + 1) 06 2r+1 = o Whence OCj = — 3 06 3 = + 5 QL 5 = — 7 a 7 = ... .% tan -1 x = a, ( x — — + - — ... ) V 3 5 / Taking x infinitesimally small, we get 0^=1 . x 3 x 5 .-. tan -1 x = x h ... 3 5 If x > 1 the series is convergent. Of this expansion the following* is a rigorous proof. Lemma — If j3 r oc-8 oc-B or-fl then tan^CX-tan- 1 ^ = sin- 1 > - >- ^(i + a 2 )(i+/3 2 ) ^(i + a 2 )(i + /3 2 ) 1+a 2 a-/3 a-/3 a— >3 and tan-^-tan- 1 ^ = tan- 1 ^- < ^- < Vo ^ i+a/3 i + a/3 i+/3 2 Now suppose that x > 1, and n is a pos' integer. x Then, since tan -1 x = tan -1 tan -1 o . 2 x . _! x + tan -1 tan 1 - n n 1 3 X -1 2X + tan -1 tan l — n n * Due to Professor Purser of Queen's College, Belfast. Expansions 317 4. -1 nx .l 1 (n — i)x + tan 1 tan -1 '— n n .-. , by applying the Lemma to each of these lines, we get that X X tan l x > ■♦©•' n ■+(")■ X X x but < — n + I+ m 2 ' n /2X\ 2 I +(— ) + .-. + X n 1 + a x ) n + ... + X n (tt •)" I + x The difference of these limits is n 1. e. is /nxv n \i + xv 1 / x 3 \ . tan _1 x differs by less than - ( ) from the sum of J n \i + x 2 / — (V*H")'- tfh (¥)'- - + 1- - where, each series goes to infinity, but n is at present 'finite. 318 Trigonometry — Chapter XVIII Collecting coeffs of like powers of x, and putting S r for i r + 2 r + 3 r + ... + (n - i) r we get that tan -1 x differs from the sum of the series x-^x 3 + ^x 5 -^x 7 + (A) n 3 n 5 n 7 v ' by a quantity which diminishes as n increases, and ultimately vanishes when n = oo. The series (A) is convergent, since it is the sum of a number of series all themselves convergent. Now hitherto n has been kept finite ; but it is a well-known algebraic result that Lim' (n = oo) ri r + 2 r + 3 r + ...+ nn ^_ v ; L n^ 1 J r + i .*. also Lim' (n = oo) - ^ Li ,r + i I r + i Hence, by increasing n indefinitely, we have, when n = oo, . X 3 X 5 X 7 tan -1 x = x 1 +... 3 5 7 where x > i, numerically. Put tan for x, and we get = tan - i tan 3 + i tan 5 0- ... where is any angle from 77" U to — 7r/4, both included. This last result is known as Gregon'e's Series. X s Note — The result tan -1 x = x + ... when used in connection with 3 Machines formula (p. 144) 4tan~ 1 ^- — tan -1 ^f ^ = 77/4 gives very rapidly converging series, from which 77 may be readily calculated. For example, in order to get 7T correct to 7 dec'l places, it is only necessary to go as far as the 7th power of \, and the 1st power of -%\-§. Factors 319 § 82. To resolve x n — 2 COS n + x~ n into factors. Since cos n + cos (n — 2) = 2 cos (n — 1) cos .-. — 2 cos n0 — — 2 cos (n — 1) . 2 cos + 2 cos (n — 2) # .-. x 11 — 2 cos n + x _n = (x - * cos 6 + x- 1 ) (x 11 - 1 + x- (n - lf ) + {x 11 - 1 - 2 cos(n -i)0 +x- (n - 1) ] 2 cos(9 - {x n ~ 2 - 2 cos(n -2)6 +x- (n - 2) ] From which, calling the function to be factorized (n), we see that 0(n) is divisible by (f>(i) if 0(n — 1) and 0(n — 2) are. But (f)(2) = (j) (1) {x + 2 cos + x -1 } /. (f)(2) and (1) are divisible by 0(i) .-. so is 0(3) Hence, by the Law of Induction, (n) is divis' by (1) Now cos n = cos (n + 2 r 7r), where r is any integer. j J + x -i But, by giving r in succession the values o, 1, 2, . . ., (n — 1), this divisor will have n distinct values : nor has it any more ; for other values of r only reproduce some of the former. .-. x n — 2 cos n + x _n = ( x — 2 cos H — jjx— 2 cos (0-\ j -j- — t jx- 2 cos(^ +2 if) + i}{x- 2cos(0 + ^) + l\ {* + (n-ovi + g 2 cos 330 Trigonometry— Chapter XVIII This may be written in either of the equivalent forms x 2n — 2 x n cos n + i = (x 2 — 2 x cos + i) -z\ — ) n L n 6 \ n / >( S7T\ 2 / 7T 2 \ 2 , — J ( i —1 where n / \ 6/ s < n /S7T\ 2 say > k I — 1 where k < 1 .\ the numerical value of each fraction in (B) < that of the corresponding fraction in the product 1 — . 2 06- sin 2 — n_ n / J sin 06 n 1 — (tF)" (C) But the sum of the fractions in (C) is absolutely convergent 111 since -g + — + -2 + . . . I 2 2 L 3 2 .'. the sum of the fractions in (B) .*. the product (B) Y 322 Trigonometry — Chapter XVIII .-. , for all very great values of n, sin 06 approximates to the product of the ist r terms of sin Oi n 06 n Oi i — ( . Oi 7T V sin — — n n Oi . 7T — sin — n n ) f . Oi sin — n a 2 7T 2 I — 27T \ n V Oi . 27T — sin — - n n j Oi" 2 2 7T 2 Proceeding to the Limit when n =00, we get / a 2 \ / a 2 \ / a 2 \ Sma = H I ~W( I ~WV I ~3v)-- When 06 is between o and 7T, sin Oi is pos', and all the dexter factors are pos'. oc 2 When Oi is between tt and 2 7T, sin 06 is neg', and 1 — — 5 7T is neg', all the other dexter factors being pos'. Sim'ly for all other values of 06. /. it was justifiable at the beginning to take the pos' sign of the root only. Cor* (1)— Put n/2 for (X, then i -i( l -?)( I -?<)( I -9')- 77 2 6 2 2 2 _ ! 4 2 - i 6 2 - 1 " 2 2 4 2 6 2 ... (Wains' formula) i-3 3-5 5-7 Corf (2)— Put 7r/6 for a, then i = 6 V ~" P/ V ~ ^TfT 2 / V ~ s 2 ?^ 2 / ' ' w - 1.86 . 144,32 4 ■ » >' — 3 5 14 3 3 2 3" Factors 323 r , . c . t a2 a 4 sin a c*^ (3) — Since 1 — — H r ... = , / a 2 \ / a 2 \ sin a and v - ?) C 1 - tt) - - -5- .". taking logs and equating, we have log - 5 + •••) - io s - 1?) + '°g - ?y - Expand and equate coeff's of (X 2 , then 3! U 2 2 2 7T 2 / TT 2 _ I I _I "6 I 2 2 2 3 2 Equate coeff's of 0C 4 -, then 6l' a\3l/" 2tt 4 Vi 4 + 2 4 + 3 4 + " ) 7T 4 I I I 90 I 4 2 4 3 4 § 84. In eq'n (A) of last § for 6 write (0 + 77-/2 n) then cos n 6 = 2 n_1 sin (0 + 77-/2 n) sin (# + 3 77-/2 n) . . . . .. sin {6 + (2 n — 1)77-/2 n} Proceeding exactly as above, there will result •;*H- 2 2 a 2 \ / 3 2 7T 2 / V 1 2 2 a 2 5 2 7T 2 n ~ sin 2 a , Or, since cos 0L = — . — , we have 2 sin 0C 2 cos (X = - 2(X «/, _ **) a -44) a - 44) a - 44) .... \ 7T 2 / \ 2 2 7T 2 / \ 3 TT / V 4 7T / / _^ / Ot 2 \ V W V 2 2 7T 2 / •*■* / 2 2 ft 2 \ / 2 2 ft 2 \ / . 2 2 CX 2 \ = V 7T 2 /V 3 8 W V 5 2 W " the other factors dividing out. Y 2 324 Trigonometry — Chapter XVIII Examples 1. To find the Limit, when m and n are indefinitely increased, but n is greater than m, of the expression (-w)('-^)-(-!)4 + 9.(' + .?-( l + S sin 77 x T . „ = . Lmr z say 77 .-. log z = log (i + ^) + ... -v log(i + jj) X X + ... + residue m +i n Residue < — ^ + ... -f — s- 2 j_(m + i) 2 n 2 J and .-. is negligible, V 2 —% is conv / t m+i m + i m+2 n — i n Now n + i m +2 m +3 n n + i _./ I + _L.yv 1 + _L_y I ...( I + i)- 1 \ m + 1/ \ m+2/ \ n/ — log = — + + ...+—] + residue °m+i [_m +i m+2 n J Residue < -^ + 7 75 + ... + — 5-, .*. negligible (m + i) 2 (m + 2) 2 n 2 ' s s 1 . n + 1 . n .*. log z = x log = x log — , ultimately m + 1 ° m ■. Limit of original product = ( — \ » x (Proof by J. Larmor) Examples 3 2 5 2. If A x A 2 ... A n zj « regular polygon of n sides; C M"} + R 2 .'., taking products of corresponding sides, the result follows by Cotes' Theorem (§82) Cor 1 (1) — Suppose P on the circumf '. n OC Then n PA„ = 2R n sin r 2 Cor' (2) — Suppose that CP produced meets O in A Then OC = 2 ir/n ... n (paJ.) = (cp n - R n ) 2 IT PA r = CP n - R n Let mj , m 2 , ... be the mid p'ts of arcs A t A 2 , A 2 A 3 , ... .-. IlPA^.IlPm^ = CP 2n -R 2n .% IT Pm r = CP n + R tt (Cotes) 326 Trigonometry — Chapter XVIII § 85. By the use of the preceding results, and by other artifices, numbers of trigonometrical series can be summed. Perhaps the most frequently useful of such artifices consists in breaking up each term of a series into two others related in some such way as the following — Let T 1 + T 2 + T 3 + ... + T n be a series of trig'l terms, formed according to some common law ; and suppose that T r = U r - U r+1 (a) where U r , U r+1 are trig'l expressions. Then the series = (U, - U a ) + (U, - U.) + ... + (U n - U n+1 ) and .-. its sum is \J X — U n+1 In the manipulation of series any relation, such as (OL) above, by which the form of the series is modified, will be called a formula of reduction. As a general rule it may be said that, where this artifice is ap- plicable, the whole difficulty of the summation consists in finding a suitable formula of reduction. Examples 1 and 2 following are done by this method. The process of breaking up T r may be sometimes facilitated by multiplying it by a suitable factor; see Example 3 below: this Example, as also its Corollaries, should be carefully studied. Example 4 gives a Theorem of fundamental importance as an aid in the summation of many series by elementary methods. To it the Student is recommended to pay particular attention. Examples 1. To sum the series cosec 2 OC + cosec 4OC + ... + cosec 2 n 0C. Here T f = cot 2 r_1 OC - cot 2 r 01 .*. the series = (cot a- cot 2 a) + (cot 2«-cot 4a) + ... + (cot 2 n-1 a-cot 2 n a) and .-. its sum is cot OC — cot 2 n OC Series 327 2. To sum the series cot- 1 foe + ^J + cot- 1 (3 a + Aj + cot- 1 ( 6a + ^) + .» . - (n (n + 1) 2 ) ... + COt- 1 1 : « + ->• I 2 CXJ Here T = tan- 1 r— -„ r 4 + r (r + 1) Or (r + i)a ra = tan- 1 (r + 1) a ra I + i £— . — 2 2 _ . . (r + i)a . ra = tan- 1 J - tan- 1 — 2 2 .-. the series = Han -1 OC — tan -1 - ) + (tan -1 tan* 1 OCJ + ... r. . (n + 1) a .nal ... + tan- 1 tan- 1 — and .-. its sum is tan -1 —. — tan -1 — 2 2 3. To sum the series sin a + sin (a + /3) + sin (a + 2 /3) +...+ sin {a + (n - i)/3}. Here T_. 2 sin — = 2 sin {a + (r - 1) B) sin — r 2 — 1 r- j 2 = cos (a + ^=- 3 /3) - cos (a + il^=i $\ series . 2 sin — = cos (a - — ) — cos ((X + — J /3--^ £)_oos( a + 4 /3\ .„/«,. 3/3 + cos (a + — j - cos (a + — j = cos 328 Trigonometry — Chapter XVII in (oc + /3 J sin n/3 sin l a + - - - a 1 sin — - l. e. the sum is . /3 sin J — 2 Cor 7 (i)— Put it + /3 for /3, then sin OC — sin (OC + /3) + sin (OC + 2 /3) — sin (OC + 3 (3) + . . . to n terms ;in j a+ ^(7r + / 3)jsin^ /3 cos — 2 C^ (2)— If /3 = 0C t we get . n + 1 . n OC sin a sin — 2 2 sina + sin2a + sin3a+... + sinna = . a sin — 2 C^ (3)— If /3 = 2 7r/n, then sin n/3/2 = sin tt = o. .*. sin OC + sin (a + 2 7r/n) + sin (a + 47r/n + ... ... + sin {a + (n — 1) 2 7r/n} = o Ctfr' (4) — In exactly the same way, or by putting 7T/2 + OC for OC, it will be found that cos a + cos (a + /3) + ... + cos {a + (n — 1)/3} / n - 1 n \ . n (3 cos la + p J sin — — sin — 2 n+i , na cos —asm — whence cos OC + cos 2 a + . . . + cos n OC = — . a sin — 2 So also cos OC + cos (OC + 2 7r/n) + . . . + cos { OC + (n — 1) 2 7r/n } = o £^ (5) — Since 8 sin 4 OC = cos 4a — 4 cos 2 a + 3 [see § 23] 8 sin 4 (a + (3) = cos 4 (a + /3) - 4 cos 2 (a + /3) + 3 Series 329 . ■. 8 [sin 4 OC + sin 4 (OC + 2 ir/n) + sin 4 (OC + 4 ir/n) + . . . ... +sin 4 {a + (n-i) 2 7r/n}] = cos 4 0C + cos 4(a + 2 7r/n) + ... + cos 4 {a + (n-i) 2ir/n} — 4 [cos 2(X + cos 2 (pc + 2ir/n) + ... + cos 2 {a + (n — 1) 2Tr/r\}~] + 3 n .-. sin 4 a + sin 4 (a + 2'7r/n) + ... + sin 4 {a+(n — 1) 27r/n} =fn Sim'ly cos 4 a + cos 4 (a + 2-7T/n) + ... + cos 4 {0( + (n-i) 27r/n} = |n The last result can be generalized thus— By an easy inductive generalization of § 23 (which the Student should work, following exactly the method of that §) it can be shown that 2 m_1 cos m = cos m0 + m cos(m-2) 6 + ... + m C f cos (m-2 r) + ... , , , . m ! the last term being — ■ — - when m is even m ! and cos when m is odd 2 2 Hence each term of the series cos m a + cos m (0C+ 2 7r/n) + ... + cos m {(X + (n - 1) 2 71/n } can be expressed as above in cosines of multiples of the angles, if m < n n m ! .. •\ the series = — • — , if m is even am (v ! ) but =0, if m is odd. i.e. cos m a + cos m (a + 2 7r/n)-K.. + cos m {(X + (n-i) 2 7r/n} is independent of OC, if m < n Sim'ly sin m a + sin m (a + 2 7r/n) + ... + sin m {a + (n-i) 2 7r/n} is independent of OC if m < n. Note — All these results should be known by heart. 33° Trigonometry— Chapter XVIII 4. To sum the series sin (a + n^-nCjSin {a + (n-i)/3} cos/3 + ... ...+ n C r sin {a + (n-r)/3} cos r /3... +sin a cos n /3, ivhere C zV //z£ number of combinations of n things, r together. Let the series be denoted by <|) (n). Also, for brevity, denote sin(a+n/3) by s n , and cos/3 by c. Then (£ (o) = s = sin a (f)(i) = s x — s c = cos a sin /3 Now sin (a + n/3) — 2 sin {a + (n — 1) /3}cos/3 = — sin a + (n — 2)/3 .*. s — 2 S n _ 1 c = — S n _ 2 {formula of reduction) .'. (j) (2) = s 2 — 2 s x c + s c 2 = — s + s c = s [-(i-c 2 )] (3) = s 3 - 3 s 2 c + 3 Si c 2 - s c 3 = (S 3 — 2 S 2 C) -(S 2 -2S 1 C)C + S 2 C 2 — S C 3 = — Sj + s c + s x c 2 — s c 3 • = (S!-S C)[-(I-C 2 )] cp (4) = s 4 — 4 s 3 c + 6 Si c 2 — 4 Sj c 3 + s c 4 = (S 4 — 2 S 3 C) — 2 (S 3 — 2 S 2 C) C + 2 (S 2 — 2 Sj C) C 2 + S C 4 = — S 2 + 2 S x C - 2 S C 2 + S C 4 = S - 2 S C 2 + S C 4 = S [-(I-C 2 )] 2 The law indicated (of which a generalized inductive proof is added) is that for the even numbers (beginning with zero) <$> ( n) is a G. P. whose 1st term is sin OC, for the odd numbers (f) (n) is a G. P. whose 1st term is cos OC sin /3 ; and the common ratio for each G. P. is — sin 2 /3 : i.e. that (|) (2 m) = (- i) m sinasin 2m £ and ( 2 m + 1) = (- i) m cos a sin 2m+1 /3* * Due to Mr. R. Chartres, of Manchester, by whom it, and several original Exercises depending on it, were kindly sent to me. — R. C. J. N. Series 331 Assume, when n is even, that (2 m + i) = ( - i) m cos OC si n 2 m+1 /3 then (2 m + 2) = (- i) m+1 sin a sin 2m+2 /3 But we have seen that the theorem is true for (j) (3) and (j) (4). .*. , by the Law of Induction, it is true for all pos' integral values of n. 33* Trigonometry — Chapter XVIII 5. To sum, both to infinity and to n terms, the series sin OC + x sin (OC + 0) + x 2 sin (OC + 2 /3) + ... -cvhere x w /&r.r ^aw unity. The series . (1 — 2X COS (3 + x 2 ) = sin a + sin (a + /3) - 2 sin OC cos/3 x J + x + sin (a + 2/3) — 2 sin (OC + /3) cos /3 + sin OC Whence we see that (using the notation of the last Example) the rth column, if r > 2, is (s r - 2 s r _ 1 c + s r _ 2 ) x r and .'. vanishes by using the formula of reduction in Example 4. sin OC — x sin (oc — 8) .'. the sum ad inf = 1 — 2 x cos /3 + x 2 But the sum to n terms will have two extra terms, owing to the last two columns not vanishing. Thus the sum to n terms _ sin OC - x sin (OC - /3) - x n sin (OC + n /3) + x* 1 " 1 " 1 sin {a + (n - 1) /3} 1 — 2 x cos /3 + x 2 6. To find the sum to infinity of the series x 2 x 3 sin a + x sin (a + /3) + — sin (OC + 2/3) + — j- sin (OC + 3/3) +... 2 . 3 . Let the sum be denoted by S : then since e we have, by multiplication of the two series, S . e -xcos P -xcose = I _ xcos ^ + i^. C os 2 /3-^- cos 3 /3 + 2 . 3- = sin a + sin (a + /3) — sin a cos /3 x + sin (OC + 2/3) -2sin(a + /3) cos/3 + sina cos 2 /3 — + sin (a + 3/3) - 3 sin (CV + 2/3) cos/3 + 3 sin (a + /3) cos 2 /3 — sin a cos 3 /3 Series 333 i. e., using the notation and results of Example 4, S . e -x cos ^ x 2 x 3 - (o) +X(j)(i) + -7 cos 2 a + cos 2 2 a cos 2 a cos 2 3 ex cos 2 2 a (14) log (1 + 2 cos a) + log (1 + 2 cos 3 a) + log (1 + 2 cos 9 a) +... (15) tan Tr/ 2 n+1 + 2 tan 7r/ 2 n + ... + 2 n ~ 2 tan tt/ 2 3 + 2 11 - 1 (16) ta n^.+tan-ij^+'^ /oA' Ctftf^: '79 (17) tan-i-l + tan-^ + tan-i^ + tan- 1 2X + --- (18) tan-ia+tan-*^^ + ^r^ + tan_1 iTl^ 2 + - Note — ^4// M (n) sin /3 16. By means of the last Exercise sum to infinity x 2 x 3 cos a + x cos (ex + /3) + — - cos (a + ?/?) + — cos (a + 3/3) + ... 2 . 3. Exercises 337 17. By means of Chartres 1 Theorem (Example 4 preceding) prove that — (1) sin n(X— nCjsin (n — 1) OC cosOf + n C 2 sin (n — 2) OC cos 2 0f — ... ± n C r sin (n-r) Ofcos r 0f . . . -n sin Of cos 11-1 OC n-i = o, if n is even; but = ( — 1) 2 sin 11 Of, if n is odd (2) sin (n + i)a- n C 1 sin n Of cosOf H- n C 2 sin (n-i)a cos 2 (X — ... ±nC r sin (n + 1 — r) Ofcos r 0f . . . + sinOf cos 11 Of n n-i = ( — i) 2 sin n+1 0f, if n vs,even\ but = (— 1) 2 cos Of sin n 0f if n is odd. 18. Also by the same Theorem sum to infinity — (1) sin Of + x sin 2 Of + x 2 sin 3 Of + ... (2) sin OC + sin (Of + /3) sin + sin (Of + 2/3) sin 2 + ... (3) sin Of + sin (Of + /3) cos /3 + sin (Of + 2/3) cos 2 /3 + ... (4) sin Of + sin 2 of cos Of + sin 3 Of cos 2 Of + ... 19. Sum to infinity, by treating as the sum of two series, each of — cos 2 Of cos 4 Of W 1 - -j|- + ~^~ - ... Note — In Exercise 16, z/"x= i,Of= o, and (3 is first 77/2 — 0, then 77/2 + 0, two series result : add these, and then write OCfor 0. (2) 2 cos Of + f cos 2 Of + -J cos 3 Of + ... 20. Following a process analogous to that in Corollary (5) of Example 3, sum to n terms sin 4 (X + sin 4 (Of + /3) + sin 4 (Of + 2 /3) + ... Note— See p. 71. 21. Show that each of the following is a particular case of Example 6 — . . . sin 2 Of sin 3 Of (1) sin Of- , + ^ ... 2 ! 3 ! sin 2 a cos 2 Of sin 3 Of cos 3 Of (2) sin Of cos Of + : + j +... 2! 3! 22. Show that each of the following is a particular case of Exercise 16 cos OC cos 2 OC M cos 3 Of cos 2 Of + : — cos 3 Of + 1 2 ! 3 (1) cos Of + — ; — cos 2 Of + — — } — cos 3 Of + — —. — COS40f +.., sin OC sin 2 Of sin 3 Of (2) COS a + COS 2 Of + : COS 3 Of + : — COS 4 Of + . . . V T 1 \ 1 ! 2 7. 33$ Trigonometry — Chapter XVI II 23. If S = x sin 6 + — - sin 2 6 + —. sin 3 6 + ... X X and C = 1 + x cos + — cos 2 0+—, cos 3 + ... 2 ! 2! u 2! prove that x 2 = (tan- 1 -^\ + {J log (S 2 + C 2 )} 2 and tan 6 = 2 tan- 1 - / log (S 2 + C 2 ) 24. With the notation, and by means of Example 4 {Chartres' Theorem) sum each of the following — (1) (f) (o) + <£) (2) + (4) + ... to infinity, and to n terms (2) 4>(°) -4>0) + 4>(4)-- (3) 0(O + *(3) + *(5)+... (4) #(0-<£(3) + (5)-... (5) 0(o) + x«^(a) + x 2 <#>( 4 ) +... , (6) ^(o)+^0( a ) + ^(4)+... , z. 4. (7) x*(i) + ^*(3) + p*(5)+..., o ■ 5 ■ (8) *(a)-i + (4) + i + (6)-». • 25. With the notation of Exercise 15 for (|) write/, in each of the fore- going, and sum the eight series produced. 26. Following exactly the process on pages 316 to 318 show that .-1 V = 1 x d 1 . 3 x 5 sm -1 x = x + - • — + — - . — +... 23 2.4 5 T . , , 1 (111 1 . -\ 1 1 ) Deduce tan~ x x = , -Ji + 5 + — - • -- T<% + —l Vi + x 2 I 2 3 i + x 2 2.4 5(i+* 2 ) 2 3 27. Show that (1 + sec a) (1 + sec 2 a) (1 + sec 4CX) ... to n factors = ■ tan 2 n_1 a tan a Exercises 339 28. Find the Limits of each of the following infinite products — (1) A - 3 tan 2 -Vi - 3 tan 2 -Y (1 - 3 tan 2 *Y... (2) (3 cot 2 j - 1) (3 cot 2 | - i) 3 (3 cot 2 f - ij... (3) A - 4 sin 2 -j (1 - 4 sin 2 -A (1 - 4 sin 2 — j ... «> 0-.2-*SH-*!)('-^S • 29. By substituting the series for sin 0, COS0, sin 2 0, cos 2 in the formulae — (1) sin 2 = sin cos (2) COS 2 = 2 cos 2 0—1 (3) cos 2 = 1 — 2 sin 2 and equating coefficients of 2n+1 in (1), of 4n in (2), and of 4n+2 in 0). three algebraic series can be summed : find them. Purkiss : E. T. iv 30. Show that — 222 2 (1) tan x = (2) TT—2X TT+2X 37T — 2X 3 7T + 2 X secx _ __i 3 5 4 7T 7T 2 — 4X 2 3 2 7T 2 — 4X 2 5 2 7T 2 — 4X 2 .flfo/A' 7W': '80 Note — •SVWtf cos (0 — x)/cos x EE cos + sin tan x ; by expanding each term of the sinister in factors, dividing and re-arranging •; and expanding sin 0, Cos z;z //z£ dexter, in powers of ; we get two series in powers of : equate coeff f s of 0. This will give (i), then apply (i) to sec x = tan (— + - j — tan \4 2/ 31. Similarly to the last find series for cot x and cosec x. Note — The expressions for tan x and cot x, in this and the preceding, come very easily by taking logs of cos x and sin X expressed in factors, and differentiating. 32. Resolve sin x + cos x into quadratic factors. Joh r CamV\ '45 Z 2 34-0 Trigonometry — Chapter XVIII 33. Prove that i + sin fl / 77 + 2d \ 2 / 3TT -1 fl V/ 5 77 + 2 \*/ >J 7T-2 fl \ 2 / 9 77 + 2 fl y/ lI77-2fl \ 2 ~ 2 \1~ A A* )\ A* A 8tt A 8tt A 12 ^r ) ' Kings Cam!/'. '90 34. If n is a positive integral power of 2, prove that /- 5z2- . 77 . 2 77 . (n — 2) 77 v n = 2 2 sin — sin — ... sin — n n 2 n ^ 7 • . X . X + 77 Note — Deduce from sinx = 2sin -sin J 22 „. X. X + 77 . X+27T. X + 3 77 = 2 3 sin - sin sin sin — = &c 4 4 4 4 35. From sin ncf) = 2 n-1 sin<|)sin(<£ + - j ... sin ((f)+ 77 j deduce — n-i . 77 . 377 . 577 . (2 n - i) 77 (1) 1 = 2 n x sin — sin — sin- — ...sin- — J 2n2n2n 2n n-l . 77 . 277 . 377 . (n — i) 77 (2) n = 2 n i sin — sin — sin — ...sm^ '— n n n n (3) tan (/) tan U> + —^ tan U> + ^J ... tan ($ + !L=_! ^ n n-l = (— i) 2 when n is even ; but = (—1) 2 tan n (p when n is odd. Note — Let the Student pay attention to these results : they will be useful in some of the Exercises follozving. 36. Prove Huygens' approximation for the length of a circular arc — Length of arc = 2/3 + \ (2 /3 — OC) v»here OC = chord of arc, /3 = chord of half the arc. x x Note — If r = rod', x = arc, then OC = 2 r sin — , 3 = 2 r sin — : 2r r 4 r expand the sines in powers of the /\ s , and neglect x 5 &c. 37. The circumference of the inner of two concentric circles (radii R. r) is divided into n equal parts by P l3 P 2 , ... P n : if A is a fixed point on the outer circumference, prove that 2 pa 2 = n (R 2 + r 2 ) Exercises 341 38. If P is a point within a square ABCD, such that II PA = XA 4 , where X is the cross of the diagonals ; and if OC denote the angle XPA ; prove that (a being a side of the square) 2 XP 4 = a 4 cos 4 (X T. C. £>. Note — Use De Moivre's Prop' given as Example 2. 39. The circumference of a circle of radius R is divided into n parts by points P 1} P 2 , ... P n , each of which subtends the same angle at a point A within the circle: if CA is a, and r l3 r 2 , ... r n are the lengths of AP 1? AP 2 , • • • AP n > prove that 2 r = (R 2 _ a 2 ) 2 - 40. A 15 A 2 , ... A n is a regular polygon; C is the centre and R is the radius of its circum-circle ; P is any point, and CP is a; PN 1? PN 2 , ... are perpendiculars on AjA^ A 2 A 3 , ..., prove that SCNjN,) = n(R 2 + a 2 ) sin 2 — n 41. A x A 2 ... A 2n+ i is a regular polygon, and P is any point in the arc A x A 2n+ i : prove that PA 1 + PA 3 + ... + PA 2n+1 = PA 2 + PA 4 + ... + PA 2n 42. If Pi> P2> '•• P2n are perpendiculars from any point in the circumfer- ence of a circle of radius r, on the sides of a regular circumscribing polygon of 2 n sides, prove that Pi Ps ••■ P 2n -l + Pa P4 ••• P 2n = rn / 2n_1 Math' Tri'\ '42 Note — Use the formula sin n + — 1 ... sin f <^> + ir) cos n (j) = 2 n_1 sin (

1 2(n + i)J 45 . C is the centre of a circle radius r ; and A is a point at which the circle subtends a right angle : from A are drawn 2 n — i secants (of which APQ is a type) and tangents are drawn at P, Q : if perpendiculars p, q are dropped from A on these tangents, prove that ,2 ri(pq) = n {r 211 - 1 ^ 11 - 1 } 5 Note — Show that IT (pq) = r 2 ^ 2n ^ sin — ... sin it and write vr ^ 2 n 2n 2 x\for n in Exercise 35, (2). 46. n points are taken at equal distances along the circumference of a circle of radius a — the first being fixed and the other varying : if p is the distance of their mean centre (for equal multiples) from the centre of the circle ; and

npsin-!— = asm( )

= o (2) The two will be identical if cos (f) = y, provided that we can choose n so that ,2 _ 3 1 — 4 and rn 3 = A cos 3 <£ qn* = i 1 i. e. if n = - \M 2 ^ q = r -J^ and /. cos 3 d> - °r 2 q3 It is .'. a necessary condition for the identity of (1) and (2) that 4 27 The method will .". apply to the irreducible case of Car dads Algebraic Solution, i. e. to the case when all three roots are real and unequal. 344 Trigonometry— Chapter XIX Let OC be the least value of 3 (f) satisfying cos 3 (p = - V^| 2 q 3 OC Then one value of y is cos — J 3 2 7T -t- OC The other values are cos = — (See p. 132) .*. the roots are /q OC /q 2TT + 0C /q 2TT — 0C 2 V v cos - , 2 V v cos , 2 V - cos — — - 3 3 3 3 3 3 Example Solve 8x 3 — 36 x 2 + 42 x — 13 = o by trigonomet?y. Put y + -| for x and the eq'n becomes y 3 -f y- i = o Writing z/n for y, we get z 3 -§ n 2 z- in 3 = Since i(-in 3 ) 2 <^ T (-f n 2 ) 3 all the roots are real and unequal, and the method applies. .-. comparing with cos 3 (J) — ^ cos (j) — ± cos 3 (f) = o we have z = cos <£, n = — =, 2 n 3 = cos 3 rf> V2 whence cos 3 0= —^ = cos - , or cos (2 tt + —) V2 4 \ ~ 4/ r 3\4/ 3\ 4/ + V3 + 1 — V3 + 1 1 ••• z = — , or — , or + 2 V2 +2 V 2 — V; + V3 + I .*. y = — , or — 1 * 2 • v - ± V 3 + 4 nr 1 . . x = — > or 2 Disjecta Membra 345 Exercises 1 . Solve the equation x 3 -6x — 4 = by trigonometry. 2. In the equations x 3 + qx — r = o, if 27 r 2 > 4q 3 , so that Cardan's method applies, the values of x are given by upper signs going together, and lower together : show that these values can be put in the forms y r sec ( cos » sina -j a. ^ 3/- / 10 • 2 v\ \/r Icoss - + sins - \ 3. Solve by trigonometry (using a table-book) the equation x 3 - 18 x 2 + 87 x - 70 = o 4. Show that, similarly to the case of a cubic, the quintic equation x 5 + px 3 + qx + r = o may be solved trigonometrically, by comparing it with cos 5 = J 6 cos 5 — 20 cos 3 + 5 cos provided that p 2 = 5 q, and that p and 4 pq 2 + 125 r 2 are both negative. Apply the method to solve x 5 + 5 x 4 — 20 x 2 — 5 x + 3 = 0. 5. AB, BC, CD are three consecutive chords of a semi-circle, whose respective lengths are as the numbers 1, 2, 3 ; find a cubic equation to give the length of the diameter AD ; and solve it by trigonometry. 6. A circle, centre C, cuts in A, B, the arms of an angle ACB : it is an easily proved geometrical fact that, if P can be found in AC produced so that, PB cutting the circle in Q, PQ is equal to CA, then the parallel through C to PQB is a trisector of ACB : show that, taking the radius as unity, and denoting CP and ACB by x and OC respectively, the equation giving x is x 3 — 3 x — 2 cos OC = o And solve it by trigonometry, in the particular case when OL is 6o°, interpret- ing the meaning of each of the three values of x. 346 Trigonometry — Chapter XIX § 87. The geometrical division of a given angle into parts whose T. F.s are in a given ratio. Let AOB be the given A, and m : n the given ratio. Case i— for sines and cosecants. With O as centre, and any radius, describe a O cutting the arms of the A in A, B. Divide AB in P, so that PA : PB = m : n Then OP divides AOB as req'd. For dropping PM, PN J_ s on OA OB, we have sin POA : sin POB = PM : PN = PA : PB by sim'r A s = m : n And cosec POB : cosec POA = m : n Case 2— for cosines and secants. Take any length OB in one arm, and OA in the other so that m : n = OB : OA Draw OP _L to AB ; then OP divides as req'd. For cos POA : cos POB = OP/OA : OP/OB ■= m : n And sec POB : sec POA = m : n Disjecta Membra 347 Case z—for tangents and cotangents. Describe any O thro' O cutting the arms in A, B. Divide A B in X, so that XA : XB = m : n Draw XD J_ to AB to meet arc AOB in D ; and let DX meet O again in P. • Then OP divides as req'd. For tan POA : tan POB = tan PDA : tan PDB = XA/XD : XB/XD = m : n And cot POB : cot POA = m : n § 88. Geometrical construction of an angle of which a T. F. is given. Example If cos 6 = v-f, construct 6 geometrically. Take CA, CB of unit length and _L to each other. Then AB = V2 Draw BD ± to BA, and of unit length. Then AD = V3 cos BAD (which = £§) - \f\ i.e. BAD = 348 Trigonometry — Chapter XIX Exercises Construct geometrically sec -1 3, sin -1 -, cosec -1 5, tan -1 ^. 2V3 See also Exercise 44, p. 150. § 89 • Some of the results ^§§32 and 33 can be easily found by geometry. Examples 1. To find sin 18 geometrically. Let ABC be a A, described by Euclid'w. 10, such that AAA B = 2 BAC = C Drop AN J_ on BC A Then BAN = 1 8° For AB put r, and for BC put x Then sin 1 8° X 2 r But X 2 = r{? - -x) (by Eud iv. 10 .-. CD 2 X + - r — 1 .'. X 2 - r + 1 = VI . sin 1 8° v 5 - 1 where + V5 must be taken, v sin 18 is pos'. 2. To find tan 15 geometrically. L N A A Let AOB be 30 . Bisect AOB by OM ; and draw AM B ± to OM. Drop BL, MN _L S on OA. Disjecta Membra 349 Then (by § 31) if BL is taken as unity, OB will be 2, and MN will be |. Put x for BM, and y for OM. x Then tan 15 = y But x : 2 = \ : y (Sim'r A s OMB, ONM) .*. xy = 1 Also x 2 + y 2 = 4 x y .-. - + - = 4 y x •: ©•-*© "- .*. tan 1 5 = 2 — V3 where — V3 must be taken, v tan 15 < 1, obviously. Exercises 1. Find geometrically sin 15 , tan 22°^, cos 36°, sec 72 . 2 . Prove geometrically that cot ii° 1 = 1 + V2 + V2 v 2 + V2 § 90. On so-called ' geometrical proofs' of some of the funda- mental formula of Chapter IV. As in §§ 18, 19 and 20, we saw that the geometrical parts of the proofs of the formulae therein treated of only applied to special cases, and required additional reasoning to make them general ; so with other fundamental formulae geometrical constructions may be used to prove them in particular cases, but such constructions do not (as a rule) prove them for other cases. Excepting in the few instances when we can exhaust the possible cases, the only way in which purely geometrical proofs can be made universal is by the generalizing principles of projection. But these principles are difficult to apply thoroughly, or to see the full force of, until considerable familiarity with the algebraic treatment of geometry by coordinates has been attained; and unless a proof by projection is carefully and thoroughly elaborated, it amounts to a mere ' De gg m g of the question.' Here follow some geometrical proofs in special cases. 35° Trigonometry— Chapter XIX 1. T. F.s of 2 (X in terms of T. F.s of OC, when 2 OC is less than a right angle. VI R K N A/ ^a) P /I - Q X Let OA, starting from coin- cidence with OX, revolve thro' 2 OC, where 2 OC < 90 . Take equal lengths OP, OQ in OA, OX respectively. Let bisector of 2 CV meet PQ in N. Drop J_ s PM, NL on OX; and NR on PM. Obviously A ONP E A ONQ so that N is mid p't of PQ, and .-. PM = 2 LN, and PR = NL. MP 2LN ON •'• s,n2a = QP= ONTOP = 2S,nacosa OM OL-RN OL ON RN PN COS20£= OP = —OP~ = ON • OP - PN ' OP = cos2a - s '" 2a MP tan 20c = =rr- M = 2LN 2LN OL 2 tana OM OL-RN 1 — RN NL 1 -tan 2 a PR " OL 2. Expansion of tan (OC — /3) when OC is within 5 of 220 , and /3 within 5° o/4° Let OA, initially along OX, revolve in pos' direc' thro' OL ; and then back in neg' direc/ thro' j3 to OB ; where OC, fl have about the values assigned. Disjecta Membra 35i In OB take any p't P; and drop J_ s , PQ on OA, PN, QM on XO pro- duced, and QR on PN. Then the sim'ly marked A s are clearly equal. MQ RP NP NP RP- MQ = MO ~ MO '"' ~ NO "~ Now ON MO + RQ 1 + MQ RQ MO *MQ , RP RQ QP v . , A And MO = MQ = QO' b y sim ' rAa .-. tan (/3, CB > CA. Draw CX equal to CA, so that X is in AB ; and drop CN _|_ on AB. CA CX Since - — = = — — = sin B sin B . the O s round CAB, CXB, have diam's of same length, D say. Then CB 2 - CA 2 = BN 2 - AN 2 = AB . BX /cb\ 2 /ca\ 2 _ ab bx •'• \d) " \d7 ~ d ' ~d~ whence sin 2 OC — sin 2 /3 = sin (OC + /3) sin (OC — (3) Next, on any line AB, equal to D, as diam' de- scribe a semi-O ; and let BAP, PAQ be OC, (3 re- spectively, where OC > (3, and P, Q are on the arc. Draw AR to meet arc, so that a PAR = /3. Let QR cut AP in X ; and make joins as in fig'. Then AQ . AR = AP . AX (See Euclid Revised, p. 287) = AP (AP - PX) = AP 2 - PQ 2 , since PQ touches O AQX. AQ AR D ^? _ /AP\ 2 /PQV * D ~\d) \d) whence cos (OC + /3) cos (OC — (3) = cos 2 OC - sin 2 /3 a a 354 Trigonometry — Chapter XIX 6. Geometrical proof of the formula of % 68. AM, AN are intern' and A extern' bisectors of A. Rest of fig' will, for brevity, be considered to explain itself. tan B -C Then Again and cot — 2 . A sin — 2 XC _ CD b - c XY " YB ~ b + c BX cos B - C b + c BX cos 2 A CX . B - C b - c CX sin . A A sin — cos — 2 2 whence (b + c) 5 -p = a = (b - c) cos B-C . B-C sin „ , . . „ A NA MD 4MX.MD Same ne / gives sin z — = -r— • = r & & 2 b c 4 be But, by Euc' ii. 8, 4 MX . MD + DX 2 = BX 2 .-. 4 MX . MD = BX 2 - DX 2 = a 2 - (b - c) 2 = 4 s 2 s 3 si n^= V S 2 S 3 be Geometrical Proofs 355 7. A circle touches the arms of an angle AOB in A, B , and OXY, any line drawn to cut the circle in X, Y, divides AOB into parts 6, (j) : to show that sin 6 sin / and .-. 00 XY 2 Ctf/— Product ±» from A, B on OXY XY 2 ^ is const' The Student can now easily prove for himself other cases of the preceding results ; and may try to find out similar proofs of other formulge. Enough of this kind of work — entirely unimportant as it is — has been now given here. A a 2 $56 Trigonometry — Chapter XIX § 91. Some generalizing methods. 1. From any known relation between sines and cosines of A 8 A, B, C ©f a A, endless other relations can be deduced thus — Let I, m, n be integers connected by the relation l + m + n = 6k — i where k may be any pos' or neg' integer. Put (X for I A + m B + n C /3„ m A + n B + I C y „ nA+IB + mC Then, since 2 k tt — OC + 2 k 7r — /3 + 2 k 7T — y = tt, .'. 2kTT — OC, 2kir — (3, 2kir — y are A s of a A. Now sin (2 k it — OC) = — sin OC and cos (2 k tt — Ot) = cos OC Hence it is at once obvious that all relations between sines and cosines of A, B, C, will hold for OC, (3, y, tf the signs of all sines are changed. 2. The plan, given on page 245, by which all formulse involving sides and A s of a A can be extended through the relations connecting the sides and A* of the pedal A with those of the original A , may be still further extended, by using the pedal A of the pedal A, and then again the pedal A of this, and so on. Thus, denoting corresponding parts of the successive pedal A s by dotted letters, we have A'=7T-2A, A"=TT-2A' = 4 A-7T, A'" = 3 7T-8 A ... a' =acosA, a"=-acosAcos2A, a w = a cos A cos 2 A cos 4 A Now A' = A . 2 II cos A (p. 245) .-. A" = — A' . 2 LT cos 2 A = — A . 2 2 II (cos A cos 2 A) A'" = — A". 2 IT cos 4 A = A . 2 3 II (cos A cos 2 A cos 4 A) And the area of the nth pedal A can be expressed in terms of the area and A* of ABC. MISCELLANEOUS EXEECISES 1. Given that — log tan 76 11' 15" = -6093 log 761875 = 5.8819 log tan 53 3' = -1237 log 233050 - 5-3675 log7T = .4971 log 180000 = 5.2553 show that, if 76 11/ 15" = radians, then x = 8 is a solution of the equation tan x = tan -1 x. Ox' 2nd Public Exam? \ '77 2. In a triangle given a, b and A, show that the product of the possible values of C is V16A 2 + (a 2 - b 2 ) 2 Draw a figure showing the ambiguity. 3. If sin a + sin (3 + sin y = o and cos a + cos /3 + cosy ::} . (X- B . B-y . y- f(fi- V 4 > is m2 (m2 - n2) - 21. ABCD is a cyclic quadrilateral, circumscribing a circle: AB, DC meet in P : BC, AD meet in Q : if p x is the in-radius of triangle PBC ; p 2 the in-radius of QCD ; p 3 the ex-radius of PAD within the angle P; p 4 the ex- radius of QAB, within the angle Q ; prove that in-radius of ABCD = fy ' p x p 2 p z p 4 . 22. If x, y, z are connected with the sides of a triangle ABC by the relation ax + by + cz = o, prove that y 2 + z 2 + 2 yz cos A z 2 + x 2 + 2 zx cos B _ x 2 + y 2 + 2 xy cos C 360 Miscellaneous Exercises 23. If P. Q, R are the respective mid points of the arcs BC, CA, AB of the circum-circle of a triangle ABC, prove that 4 area A PQR = 2 area hexagon PCQARB = area principal ex-central A 24. Two circles are drawn through the circum-centre of a triangle to touch the same two sides ; if p 1 , p 2 are their radii, prove that 11 r _2R + r Pi P2 ~ PlP2~ R 2 Tucker'. E. T. xvi 25. Prove that 2 2 sin 2 (A-B) + 2 2 sin (A-B) sin (C-A) cos A = (1-8 n cos A) 2 sin 2 A 26. If x cota + y cot/3 + z coty = (x + y + z) cot a cot/3 coty and (x + y) cot OC cot /3 + (y + z) cot /3 cot y+(z + x) cot y cot prove that 2 x sin 2 OC = o and 2 x cos 2 OC = Queen's Camt/: '48 Dty I a = o ) a = o j a = o ) 27. H x, = sin(g : a) , * - sin ^ sin (a-y) sin (a-/3) ' sin (/3-a) sin (/3-y) ' sin(0-y) x 3 = — — — : , and S_ = x; + x^ + x£ , 3 sin(y-)8)sin(y-a) r l 2 3 prove that — = — • — 725 Note — Show that x 1} x 2 , x 3 are roots of x 3 + p 2 x + p 3 = o ; and deduce from Newton's relations between roots and coeff's. 28. C is the centre of a circle; AC a radius produced through C to B, so that CB is one-third CA; P is any point on the circumference ; PB produced meets the circle in Q ; QC produced meets PA in R : if-0 is the angle CRA, and (j) the angle CAR, prove that 1 — sin 6 /i — sin ^)\3 1 + sin 6 \i + sin <£/ W. Roberts : E. T. 29. If m sin (9 + (f>) = cos (d — (j>) prove that 1 12 1 — m sin 26 1 — m sin 2 , tan <£' so that is maximum. 34. If /, . COS COS — COS (D cos - 2 ^ 2 oo.(*-4) + co.(*-f)"' prove that cos + cos (/) = 1 35. If OC, (3, y are the three values of 0, unequal, positive and each less than two right angles, satisfying the equation a tan 3 + b tan + c = o prove that sin (OC + /3 + y) = 2 sin OC sin /3 sin y } and 2 (a + b) tan (CX + /3 + y) + c = o ) Jes' , Christ's and Emm' ', Caml/: '90 36. P is any point in the plane of a triangle ABC : PX, PY, PZ are iso- clinals to the sides, making with them the same angle )- 39. If tan (0 - <£) = cos OC tan $, show that (1 - sin 2 a sin 2 <£) (1 - tan 4 - sin 2 0) may be expressed as a perfect square. .. Tf _ cosd) (4 — sin 2 d>) 40. If cos = — show that the value of 4 + 5 sin 2

— v tan in terms of 0. 2 (1 + m) tan 2 = f , find 1 — m tan 2

COS COS 2 2 2 2 . oc + 3 . y + b oc-y 3 -b sin - — sin - cos cos cos 2 OC + y P-y P+b cos — — - + cos COS 2 OL-y 0L + 3 y + b oc- y 3-b cos cos COS ' cos — where OC, (3, y, b are respectively the internal angles made by AC with AB, BC, CD, DA. Heppd\ E. T. lvi 45. In a triangle ABC, prove that s\n i — 2 COS 2 — 2 • 2 B sin 2 — 2 2 B cos 2 — 2 sin' cos^ 2 c (a + b + c) (a - b) (b - c) (c - a) 2 abc Pet? Caml/ : 'go 46. If #i + 6 2 + 0$ + 0L 1 + 0C 2 + 0C 3 = o, prove that tan (0 X + a x ) tan (0 a + 0C X ) tan (0 3 + a x ) tan (0 X + a 2 ) tan (0 2 + 0C 2 ) tan (0 S + a 2 ) tan (0 X + a 3 j tan (0 2 + a s ) tan (0 3 + 2 Q = ba 2 ab 2 be 2 ca cos A cos B cos C where the letters are parts of a triangle ABC, prove that P = Q2a 2 ac ab be be cos B cos C 3 6 4 Miscellaneous Exercises 48. Prove that the evaluation of each of the following determinants leads to the given result — (0 II II i i cosy cos/3 i cos y i cos Oi i cos/3 cos a i = - i6TIsin 2 - 2 (2) III sina sin/3 siny cos a cos /3 cosy (3) Same determinant = — 4 IT sin = 2sin(a-/3) /3-y (4) /3-y y-a Oi- /3 cos cos cos 2 2 2 /3 + y y + Oi CX + /3 cos sin cos cos 222 /3 + y . y + Oi .a + /3 sin sin = either of last results Note — Multiply the results of (2) and (3) giving the square of the deter- minant. (5) . oC B o sin^ — sin^ — 2 2 A sin* o sin 2 _ . o B . o A sin 2 — sin 2 — o 2 2 where A, B, C are angles of a triangle. = (2 sin A) 2 Baltzer ANSWEES Note — Many of the results following are approximate: unless otherwise stated, 3-1416 is taken as the numerical value qfir. Page 15 1. i3°-846i53 2. '(-22689) 3. (1-6534) 4. tt/i2 5. 180 n Oc/ttP 6. 1 ft 6^ in 7. 143 14' 2o"-8o9 8. 6875.4 yds 9. loin, 14.142 in, 17-328 in 10. 3°°» 2 5°> I2 5°J or 30 , 15 x 100', 45 x 10000" 11. 1-15192 miles Page 16 12. 57'- 2 9578 13. 57' 2 9578 ft 15. 180/(180 - ir) = i°-oi9 16. 3-15 17. -307 1 20. 3-1416 23- -i62sqft Page 17 24. -0905 25. 185575 miles per sec 26. 2276 miles per hr 27. 1 108000000 grs 28. 26590439766400000 using 3.1415926 26590502400000000 using 3-1416 error in excess 62633600000 sq miles 29. N° sides are f? or or or 8j 12J 32 J 72 J IO 7T 9TT l800 7T 30. 31. 1819 1819' 1819 2250 IT 2500 TT I539 7T 6289 ' 6289 ' 6289 366 Answers Page 18 35. Same as 29. „„ 2omn - 18 20 mn — 18 , m = ist ratio 36. , — ; v ; where 10 n — 9 m(ion — 9) n = 2nd ,, 37. 3° 2' 27", 2 44' 12", i74°i 3 '2i" 38. 4 , 16 39. d 2 (5 77 - 6 Vti/24 2X(X+l) 2X+I Page 31 1. 2 X 2 + 2 X + I 2X 2 + 2X+ I 2. sm a = — — = cos a V2 3. ± q/Vp 2 + q 2 , ± p/^p 2 + q 2 4. 1 5. cos a = -999, tan a = .012, cota= 83.321, seca = i-ooi, coseca = 83.333 6, 3 V2 + 2 8. sin a = V2 vers a — vers 2 a Page 51 l.o 2. 1 3. (1 - sin a) (1 + cos a) 4. o Page 55 1. 1 2. - sin 3 - cos 13 Page 74 1. + + 2. - - 3. 2 n 77 ± 77/4 4. 2 n 77 + 3 77/4, 2 n 77 + 5 77/4 Page 85 1. (1) -3535, '169 (2) .612, 1.959, -1.379 (~\ 527 11.75 3 _ h ftr l \5) 6"2 53 102 965 / U1 7 12 , 2035 . S28 ( 4 ) _JL_ or ±^. + ^ ^ ; 12 5 ' — 2 1 or + 9 7 — 219 7 , . 20 V6 V6 + 1 //r . 2mn Page86 (5) ± — , ± — -j- (6) ± ^^ „ „ . a 2 + b 2 - c 2 . c 2 + a 2 - b 2 Page 90 15. sinx = ^ , s.ny = — c 2 -b 2 sin0 = — ,, a V2 b 2 + 2 c 2 - a 2 Answers $6y Page 91 21. tan (a + j3) 22. 1 + Vi — m 2 m 2 — 1 + Vi - m 2 23. - |, - f 28 (1) ( a2 + b2 ) 2 ~4b 2 (a 2 + b 2 ) 2 - 4 a 2 **' ^ 4 (a 2 + b 2 ) {2) 4 (a 2 + b 2 ) " (3) , *»? ,„ (4) 4a (5) (a 2 + b 2 ) 2 -4a 2 w a 2 + b 2 + 2b (b 2 -a 2 )(a 2 + b 2 -2) a 2 + b 2 ,*x u( u<2 ( a2+ b 2 ) 2 - 4 a 2 ) 1 */ "1 I Page 100 1. + -, + — - 2. ± 1, ± 2 3. 1 + — = 2 2 - ^3 Page 102 1. o 3. V2 v 2 + V2 — V2 — 1 5. V6 + V3 - V2 - 2, V6 — V3 - ^2 + 2 Page 103 10. PXC = 67 J, PCX = 22 J Page 107 2. 32 x 5 — 16 x 4 — 32 x 3 + 12 x 2 + 6 x — 1 = o 3. 8x 3 + 4X 2 — 4X — 1 = o 4. i28x 7 -64X 5 — i92x 5 + 8ox 4 + 8ox 3 — 24X 2 — 8x+ 1 = o Page 126 4. n 77 ± 77/4 5. 2 n 77 + OL 6. a n it + OL 7. No Page 127 8. (1) 2 n 77 ± 71/3 (2) 2 n 7T + 77/3 (3) n 7r + 77/3 (4) 3 O—irfa - 2 n 77 + 77/3 (5) n 77/2, 2 n 77 + 2 77/3 (6) (4 n + 1) 77/2, (6 n + 1) 77/3 (n - 1) (7) n 77 368 Answers (8) n 77, n 77/2 + (— i) n 77/4 (9) n 77 ± 77/6 (10) n 77, 2 n 77 + 2 77/3 (11) n 77 + 77/4 (12) n 77 + 77/4 (13) 2 n 77 + 77/5, 2 n 77 + 3 77/5 (14) 2 n 77 ± 77/6 (15) (4 n + 3) 77/4, (4 n + 3) ^/ 8 (16) (4 n + 1)77/8, (411-1)77/4 (17) (4 n + 1) 77/8, - (4 n - j.) 77/4, 77/4 (18) 2 n 77 + 77/6 (19) (2 n + 1) 77/5, (2 n + 1) 77/7 (20) n 77/2 + OC, n 77/2 (21) (4 n - 1) 77/6 + OC Page 128 (22) n 77 ±77/3 (23) n 77 + 3 77/4 (24) 2 n 77 + 77/4 (25) 2 n 77 ± 77/2 or cos = ~ — — 56 + Y s 12. 3, J; rri77, 11177177/4 15. (611 + 1)77/3, (2 n + 1) 77 (26) (2 n + 1) 77/41 a (27) = n 77 or tan = + V f Page 129 16. 15°° 17. 70 , tio°, 190 , 230 , 310 , 350 20. n 77 + 77/4, n 77 + 77/6 21. ± Vz, + — — ' V3 Page 130 25. (1) n 77 + (- i) n 77/6 (2) n77 + 77/3 Page 134 2. tana, tan (a + 77/3); x 3 -3x 2 tan 3 OC-zx + tan 3 a =0 5. cot a + cot (a + 77/3) + cot (pc - 77/3) = 3 cot 3 a cot oc cot (a + 77/3) + cot (oc + 77/3) cot(a - 77/3) + cot a cot (a -77/3) = - 3 cot oc cot (a + 77/3) cot (a + 2 77/3) = cot 3 a 6. 6 Page 13$ 8. ± tan a, tan (77/4 ± OC) 9. 12 Answers 369 10. x 5 tan 6 0i + 6x 5 - 15 x 4 tan 60c- 20 x 3 + I5x 2 tan6a + 6x-tan6a = o 15. x 5 -55x 4 + 33ox 3 -462x 2 + i65x- n = o Page 146 32. Out of place here: compare the fig' on p. 279, with the results at end of p. 277. Page 147 37. (1) ± J (2) V 2 cot 77/12 (3) i+\/ 2 a/(i + a) (4) + h \ (5) V3 (6) + (V3 + 1) 38. (i)n7T + - or nw ± icos- 1 -^- — ' 4 8 (2) 2 n 77 ± COS -1 — =— o . . . . sin (a + 8) (3) n 77 + tan- 1 . K . \X v/ sin a sin /3 (4) 2 n 77 ± cos -1 — — or 2 n 77 + cos -1 ^5 5^i3 , - . 2 sin asm /3 (*) 2 = n 77 — tan -1 — — - ^~ voy sin (a + /3) /^ ^ , ^n • 1 sin 2 a (6) n 77 — a — (-1) sin -1 (7) 2 6 - a = tan- 1 f m " " tan oc) w/ \m + n / n , 2 sin a sin /3 Page 148 (9) 2 = n 77 + tan- 1 sin(a + /3) (10) cosfl= . q , COS(/>= V^" Vna - d 2 q pq - p B b 37° Answers (ii) Add, subtract, and multiply results; this gives cos 6 /cos (f) : then by division tan 0/tan $ is found. The rest is obvious. (12) + 4> = m7T + (- i^sin-Ha + b), _ <£ = n 77 + (- i) n sin- 1 (a - b) (13) 6 = nr + tan- 1 3^ .(+ 2V / n 2 -3m 2 -n)t £n 2 — 4m 2 } (14) 2 = sin- 1 r 3 Q + sin- 1 J^ | 2 = sin- 1 ^ -sin- 1 ^) ^ 20 = sin- 1 gV + sin- 1 T V| 1 2( |) = sin- 1 3 1 o -sin- 1 ^ ) , ,, . . 1 a 2 f/3 2 -2 . , . .a 2 -^ 2 (15) + 4> = sm ^ » 0- = sm - 1 ^ +T ^2 (16) / j j. -y — (18) 2 n 77 + sec -1 (tana tan (3 ± sec a sec/3) (19) n 77 + 77/12, = n 77 + i cos -1 ( ) ' \5-4vV (20) 2 n 77 + 77/3, 2 n 77 + cos -1 \, n 77 + tan -1 2 Page 150 46. 2 R rsin OiJVR 2 + r 2 + 2 R rcosa Page 158 1. (1) (a 2 - b 2 ) 2 = 16 b (2) a + b + 1 = ab (3) (ab)»(a»+ b»)= 1 (4) (a 2 + b 2 ) 2 - 3 (a 2 + b 2 ) - 2 b = o Answers 371 2 1 2. (5) a 3 — b 3 = 2 » (6) cos (a - /3) = (aa' + bb')/(ab' + a'b) Page 159 (7) 2 (a 2 + b 2 ) 3 = (a 2 - b 2 ) 2 (8) cos (a - (3) + sin OC sin /3 = o (9) sin a /sin /3 = (1 + m)/(i — m) (10) sin a tan z + sin /3 tan x = sin (OC + /3) tan y (11) (a 2 — b 2 — c 2 ) sin OC = 2 be (12) (a 2 + b 2 -c 2 ) 3 = 4 a 2 b 2 (a 2 + b 2 -c 2 ) + a 2 b 2 c 2 (13) {2(x 2 + y 2 ) 2 -3ax + a 2 } 2 = a 2 {(x-a) 2 + y 2 } (14) a/(b - c) + b/(c - a) + d/(a - b) = o oc — 3 (15) cos = + ksin (a + ^3) 2 (16) 2 m = cos a ± cos 2 /3 (17) (a 2 + b 2 ) b 2 = 4 c 2 (a 2 + b 2 - c 2 ) (2 a 2 cos a) — 2xy) 2 (18) (2 a 2 cos to — 2 xy) 2 + (x 2 — y 2 ) 2 _ {x 2 + y 2 -a 2 (i + cos 2 a))V (2 a 2 — x 2 — y 2 ) 2 cos 2 co tan 2 a cos y (cos /3 — cos a) Page 160 (2) — — = tan 2 y cos a (cos p — cos y) (3) (a + b)(m + n) = 2 mn (4) a 2 - c 2 = m (b 2 - c 2 ) (5) (a - b) {'(a + b) 2 - c 2 } + 4abc m = o (6) 2 ab = c (b 2 - 1) (7) (a 2 - b 2 ) 2 = 2 c 2 (a 2 + b 2 ) (8) a + b + 2 ex + 2 dy = bx 2 + ay 2 (9) m 4 + n 4 = J b b 2 37 2 Answers Page 161 3. (x/a) 2 + (y/b) 2 = i 4. (ax)* + (by)* = (a 2 -b 2 )* 5. A*x*/a* + B* y^/b* = (A 2 - B 2 )* Page 162 9. y 2 = ax 10. b 2 = (a + c) 2 , cos0 = - b/(a + c) = ± 1 12. (a - b) x 2 = b 2 (a + b - 2 c) 13. (x/a) 2 + 2 xy Vfj? - i/(ab/m) + (y/b) 2 = 2 Pa g e:6 3 17.(A.-i)(B-±)(B-^) + c(^ + i-±) = BC 2 Page 164 20. {a 2 b 2 -(a 2 -b 2 )c 2 } sin x siny + {a 2 b 2 + (a 2 -b 2 ) c 2 } cosxcosy = (a 2 + b 2 ) c 2 - a 2 b 2 2 2 2 22. (x + y)* + (x — y)* = 2 a* 23. (x/a + y/b)* + (x/a -- y/b)* = 2 24. a 2 — b 2 — c 2 + 2 ac + 4 be = o Page 165 25. abm 3 + b (a + 1) m 2 + (b + c) m + be — ad = o 2 2 ~ 2 26. (xcosa + ysina)*+ (xsin a — ycosa)*= (2 a)* 28. p 2 — ap cos Oi — 2 a 2 = o 30. p 2 — 2 pq coso) + q 2 = sin 2 co where co = 3a-2/3, P = 4 (^) - 2>(^)> q = i-2^j y V 2 xy cos 2 A Page 166 31. (-^-fX+ (-^-n) 2 - \cos B/ \cos C/ cos B cos C -4RsinA(— ^-= + ^-^) +4R 2 sin 2 A = o T \cos B cos C/ Answers 373 34. {c 3 + c 2 -2(a 2 + b 2 )} 2 {(c + i) 2 - 4 (a 2 + b 2 )} = 16 a 2 b 2 (2C + 1) 3 35. By addition and subtraction, we get sin 3 6 = k cos 6 and cos 3 6 = W sin 6 where 6 = (J) + tt/i2, k = OC (V3 — i)/(x + y), k'=a(V3-i)/(x-y) Page 180 2. (1) o (2) 1 Page 181 3. (1) 77/648000 (2) 61 77/648000 (3) a//3 ( 4 ) (a// 3 ) 2 4. 2 8. 1-0355 12. -a 2 /(2/3 2 ) Page 182 15. (1) i (2) 1 (3) (5) - 5V (6) A ' (8)8(^) 2 (9) T2 Page 190 1. .6073257 2. 34° 29' 45"- 7 3. -5003233 4. 6o° i' 22"-6 (4) n \iJ 12 5 (10) #i Page 193 1. 9-7299388 2. 9-59 o6 3 6 3 3. 9-5050321 4. 2 2°28'i6".43 5. 20 35' 15^.95 6. 44° 59' 3o"-7 Page 199 3. Within (2^)" Page 216 13. 28 feet Page 219 41. 3. 4> 5> 6 374 Answers Page 224 1. (1) 2 sec a sec /3 sin ^-^ sin @ 22 f.) - 4 sinfsinf cos (A + 5^) = sin (p (3) i°, 2 tan 6 sin ?-^ sin - — * 2 2 where b = a cos v 2°, a sin (0 + (p) = bsin Page 226 10. 2 sec sin -1 12.cosx = - CosAcos y + B ), COS(p where tan y = tan ^(2 m - n) — + — > ( 3 4) --, n7r.nA.nB.nC 16. —4 cos — sin — -sin — sin — , 2 2 2 2 when n is 4 m, or 4 m + 2 . n tt nA nB nC 4 sin — cos — cos — cos — , 2 2 2 2 when nis4m + ior4m + 3 19. p 2 = i ( a a + b 2 ), tan * = tan X ~ y + ** ~ ^ Page 361 31. (A 2 - B 2 )/ 4 A 33. tan$ = V//, tan ' =-^ Page 362 37. ± 1, ± (m 2 - 3)^(2 m) 38. + sin (0 + ) 39. {cos (2 - 0) cos ) } 2 40. j 41. + msin0 cos0/Vi - m 2 sin 2 42. (sin 0! + sin $ 2 ) 2 /(sin 2 ^ - sin 2 > » V k >'> > I > k > »)>>>*>