im Hollinger Corp. pH 8.5 LC 6301 .U5 P55 Copy 1 j/ ZlniversitY and School Extensien PLANE TRIGONOMETRY. A. W. PHILLIPS, 1889. Yale University. le g ^ 01 \ Copyright, 1889, By A. W. PHILLIPS. V Press of J. J. Little & Co. : Astor Place, New York. Course in Trigonometry. OBJECT OF THE COURSE. § i. The Plane Trigonometry is intended to teach the student the derivation of the formulas and the methods of solving, both with and without the use of Logarithms, all the cases of right- angled and oblique-angled triangles, together with the application of these methods to the measurement of heights and distances, to field surveying with the compass and theodolite, and also to the simpler problems of" Navigation. The Spherical Trigonometry covers the derivation of the formulas and the methods of solving all the cases of right-angled and oblique-angled spherical triangles, with their applications to the simple problems of the terrestrial and celestial globe. METHODS OF STUDY. § 2. The student may make use of any of the standard element- ary text-books used, in the academies and colleges, and can master topic after topic in the order indicated in the syllabus, working the problems from the text-book. It is strongly recommended, where it is possible, to use the compass and theodolite or even more roughly constructed instru- ments for measuring angles, and to work problems constructed from the student's own measurements. UNIVERSITY EXTENSION. THE DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS. § 3. The definitions of the Trigonometric functions, viz.: sines, cosines, tangents, etc., are given either as certain lines drawn in or about a circle of unit radius, or as certain ratios of the sides of a so-called triangle of reference. The advantage of the first method is that the student is always able to represent these functions by definite lines in a figure, which gives his problems more of a reality than the second method. In the method of ratios which is more generally adopted in the recent works published in this country, the early conceptions are much more difficult ; but since these definitions are the formulas of right triangles the student is greatly aided in becoming familiar with the fundamental principles in the solution of triangles. The following figures will furnish an example of each of the several functions in the four quadrants : A' JL' Fig. 1. 1st Quadrant. Fig. 2. 2d Quadrant. COURSE IN TRIGONOMETRY. Fig. 3. 3d Quadrant. Fig. 4. 4th Quadrant. For the angle AOBor the arc A B in each figure. B E is the sine, O D is the secant, O E the cosine, O C the cosecant, A D the tangent. A' C the cotangent. RULE OF SIGNS. For all -vertical lines, sines and tangents : Those above the horizontal diameters are + . Those below, — . For all horizontal lines, cosines and cotangents : Those on the right of the vertical diameter are + . Those on the left, — . For oblique lines, secant and cosecant : Those reckoned from the centre towards the extremity of the arc are + . UNIVERSITY EXTENSION. Those reckoned from the centre in the direction opposite the extremity of the arc are — . [The radius is always to be regarded as positive.] TABLE SHOWING THE SIGNS IN EACH QUADRANT. ISt. 2d. • 3d- 4th. Sine + + - - Cosine + - - + Tangent + - + - Cotangent .... + - + - Secant + - - + Cosecant + + - After learning the definitions of the Trigonometric functions, and the signs of these functions in the several quadrants, the next step is to derive the following FUNDAMENTAL FORMULAS FROM A FIGURE. (i) sec Oi COS X (2) cosec x == sin x COURSE IN TRIGONOMETRY. Sill X COS X (3) tan x — (4) cot x = cos x sin x 1 (5) tan x — (6) sin 2 x + cos 2 x — 1. cot x (7) sec 2 x = 1 + tan 2 x. (8) cosec 2 .# = 1 + cot 2 x. VALUES OF FUNCTIONS OF CERTAIN ANGLES § 5. VERY FREQUENTLY USED. sin == cos = 1. sin 30 = 1 iTj , cos 30 — 2 2 sin 45 = V~T VJ , cos 45 — 2 2 sin 6o° = VJ 1 /■ _ 1. Draw figures and show < sin 45 2 2 sin 90 = 1 ; cos 90 — o. 2. From these values of the sin and cos, derive values of the tan, cot, sec, cosec of o°, 30 , 45 , 6o°, 90 . 3. From these values derive, by the aid of a figure, the sin, cos, tan, cot, sec, and cosec, of 120 , 135°, 150 , 180 , 210 , 225 , 240 , 270 , 300 , 330 , 360 . UNIVERSITY EXTENSION. GRAPHICAL SOLUTION OF TRIANGLES. § 6. The parts of the triangle which are given may be laid off —the sides on any convenient scale, and the angles by means of a protractor — and the remaining sides and angles may then be measured. The results will give a good approximation if the meas- urements are carefully made — and will serve also as a check on the calculations. SOLUTION OF RIGHT TRIANGLES. § 7. Let A, B, C be a right triangle. A the right angle, and B the acute angle at the base. Call the sides opposite A, B, and C respectively #, b, c. b opposite side sin B = — or sin of an angle = a hypotenuse c adjacent side cos B = — or cos of an angle — — a hypotenuse b opposite side tan B = — or tan of an angle — c adjacent side #2 = ^2_|-^2 or square of hypotenuse = sum of squares of the other two sides. COURSE IN TRIGONOMETRY. APPLICATIONS. § 8. To problems in heights and distances where the triangles formed are right-angled. To problems in field surveying with the compass. To problems in Navigation in Plane Sailing, Middle Latitude Sailing, and Mercator's Sailing. EXAMPLES. Find the remaining parts in each of the following right trian- gles. Given : Base 60 feet, angle at the base 43 °. Base 75 feet, angle at vertex 76 18' 12". Perpendicular 62 feet, angle at the base 47 18'. Hypotenuse 87 feet, angle at vertex 8$° 10'. Hypotenuse 50 feet, base 37 ft. Find the distance across a stream, or to a given inaccessible object, by constructing a right-angled triangle and measuring one side and one angle. 7. Find the height of a tree, a steeple, or any object, by measur- ing the base of a triangle, and the angle of elevation of the top of the tree. UNIVERSITY EXTENSION. 8. Find the contents of a field by dividing it up into right-angled triangles and measuring such portions as are necessary in order to find the area. 9. Survey a field with a compass and chain, and explain the method of finding its contents. 10. In Plane Sailing, show how to find : (a) The departure, given the bearing and distance. (b) The difference of latitude, given the departure and dis- tance. (c) The distance, given the beaming and difference of latitude. [Draw a triangle and label the sides and the angle properly.] 11. In Parallel Sailing, draw a right triangle and label its parts so as to show the relation between the distance measured on a parallel, the latitude of the parallel, and the radius of the earth. 12. In Mercators Sailing, draw a right triangle and label it so as to show the relation between the bearing, the distance sailed, the departure, the difference of latitude, the difference of longitude, and the meridianal parts. 13. Given the latitude and longitude of two places, find the bearing, the departure, and the distance sailed in going from one to the other. COURSE IN TRIGONOMETRY. SOLUTION OF OBLIQUE-ANGLED TRIANGLES WITHOUT LOGARITHMS. § 9. Let A, B, C be an oblique-angled triangle, of which A B is the base, and the angle B either acute or obtuse. The sides opposite A, B, and C are respectively a, b, and c. Let fall the perpendicular C D on A B or A B produced. a sin A Then b sin A = a sin B ; or — = (A.) b sin B Making in turn a and b the base, we obtain similar expressions b c for — and — . We therefore have the Rule: The sides of a triangle c a are proportional to the sines of the opposite angles. Again, we have from the triangle a 2 = c 2 + b 2 — 2 be cos A (B.) Making in turn a and b the base, we obtain similar expressions for b 2 and c 2 . We therefore have the Rule : The square of the side opposite any angle of a triangle is equal to the sum of the squares of the other two sides diminished by twice the product of these two sides into the cosine of the opposite angle. CASES. Two sides and an opposite angle Two angles and an opposite side Two sides and included angle Three sides Formula (A). Formula (B). 12 UNIVERSITY EXTENSION. DEVELOPMENT OF FORMULAS. § 10. The formulas for the sine and cosine of the sum and dif- ference of two angles may be derived directly from the following figures: In Fig. 5, call the angle A O B or arc A B x, and B O C or B C y ; then A O C or A C is x + y. O K is the cos y ; C K is perpen- dicular to O B, and is the sin 7 ; C D and K H are perpendicular to O A ; K L is perpendicular to C D. The angle KCL = AOB = x CD^KH+CL C L = t) K 90s x; = sin y cos x. K H = O K sin x = sin x cos y. C D = sin (x + y) = sin x cosy + cos x smy. O D = O H - L K. O H = O K cos x = cos x cosy. L K = CK sin y = sin x sin y. O D — cos (x + y) = cos x cosy — sin x siny. COURSE IN TRIGONOMETRY. 1 3 In Fig. 6, call angle A O B or arc A .B x, and COBorCB;. The angle CHF = i,CHis the sin of y, and O H the cosine. H D 'is perpendicular to O A, and C F to H D. CE = HD-HF. H D = O H sin x = sin x cos y. HF =HC cos x = cos x sin y. C E = sin (x — y) — sin x cos y — cos x sin y. OE = OD + FC. OD = OH cos x = cos x cos y. F C = H C sin y = sin x siny. OE = cos (x — y) = cos x cosy -f sin x siny. Collecting these formulas together, we have : Sin (x + y) = sin x cosy + cos x siny. (i.) Sin (x — y) = sin x cos y — cos x sin y. (2.) Cos {x + y) = cos^: cosjy — sin x siny. (3.) Cos (x — y) = cos x cosy 4- sin jp sin jy. (4.) Putting j = x equation (1) becomes : Sin 2 x = 2 sin x cos x. (5.) Equation (3) also becomes : Cos 2 x = Cos 2 .* — sin 2 #, or cos 2^ = 2 cos 2 x — 1, or cos 2 .# = 1 — 2 sin 2 .#. (6.) Adding (1) and (2) : Sin (x + y) + sin (.* — jy) = 2 sin 3; cos>». (7.) 14 UNIVERSITY EXTENSION. Subtracting (2) from (1) : Sin (x + y) — sin (x — y) = 2 cos x sin y. (8.) Adding (3) and (4) : Cos (x + y) + cos (x — ■ y) = 2 cos x cosy. (9.) Subtracting (3) from (4) : Gos (x — y) — cos (x -\- y) = 2 sin x sin y. (10.) Putting x -f- y = A, and o: — y = E, (7), (8), (9), (10), become A + B A - B. Sin A + sin B = 2 sin cos (n-) 2 2 A -f B A - B. Sin A — sin B = 2 sin cos ( I2 -) 2 2 A + B A - B. Cos A + cos B = 2 cos ! cos (13.) A + B A - B. Cos B — cos A = 2 sin sin ( I 4-) EXERCISES. -n • t/ A / 1 cos x i. rrove sm % x = A / x=j/i. T/ x + COS X 2. rrove cos V COURSE IN TRIGONOMETRY 3. Derive sin 3 x 4. Derive cos 3 ^ tan x + tan y 5. Prove tan (x + y) = 1 — tan x tan y tan jc — tan y 1 4- tan x tan y cot a- cot y — 1 cot y + cot x cot x cot y + 1 cot y — cot x 6. Prove tan (x — y) = 7. Prove cot (x + y) = 8. Prove cot (a* — ^) = 9. Show that Formulas (1), (2), (3), (4) are true when either x or y or both are greater than 90 10. Show that sin (90 + x) = cos x 11. Show that cos (90 + X) = sin .r 12. Show that tan (90 + x) = cot x 13. Derive sine and cosine of 15 , 75 -Y*r 14. Show sine of 18 15. Given sin x + cos x — 1 : find x l6 UNIVERSITY EXTENSION. SOLUTION OF OBLIQUE-ANGLED TRIANGLES BY THE USE OF LOGARITHMS. § ii. The tables of logarithmic sines, cosines, tangents, and cotangents are the logarithms of the numbers in the natural tables of logarithmic functions — the characteristic in each case being increased by 10, to avoid the use of negative characteristics. The text-books show how to use these logarithms. Since it would be a great inconvenience to use logarithms in formulas where the terms are connected by the signs plus and minus, it is necessary, in order to adapt formulas (A) and (B), oblique- angled triangles, to logarithmic computation, to transform them so the terms shall be products or quotients. The foregoing chapter in Trigonometric analysis furnishes the means for making these transformations, and also forms the basis for solving many problems not necessary in the solution of triangles. a sin A Formula (A) : — = by composition and division becomes b sin B a 4- b sin A + sin B sin A — sin B Dividing (n) by (12) and reducing: sin A + sin B tan X A (A + B) sin A — sin B tan X A (A — B) COURSE IN TRIGONOMETRY. 1 7 Whence a + b tan % (A + B) - (C) a - b tan ^ (A — B) Formula (B) : a 2 = e 2 + b 2 — 2 be cos A, may be written : cos A = 2 be From (6) : e 2 + b 2 - a 2 (£ + c + a) (b + c - a) 2 cos 2 y 2 A= 4- 1 = 2 &■ 2 be (IS-) From (6) : c 2 + £ 2 - a 2 (0— * + f) (0 + .4 - * ) 2 sin 2 ^ A == 1 — ■ = 2 be 2 be (.6.) # -f b + <: Putting -= s, and dividing (16) by (15) and extracting 2 the square root : s (s—a (s-a) (D.) SUMMARY. a sin A 12. - = (A.) b sin B UNIVERSITY EXTENSION. a + b tan y 2 (A + B) = (C) a - b tan ^ (A - B) a / {s - c) (s - b) tan y 2 A = \/ -± M ^ (D.) A + B + C = 180 . (E.) CASES. — Formulas for complete solution. Two sides and opposite angle. (A), (E). Two angles and opposite side. (A), (E). Two sides and included angle. (E), (C), (A). Two angles and included side. (E), (A). Three sides. (D), (A), (E). In the above formulas and also in the . cY A formulas of oblique-angled spherical triangles, ^ a the formulas will hold true if, in place of each rk letter, we substitute the one next in order, taken c b C" B in the direction of the arrow in the figure annexed. This amounts to the same thing as turning the triangle, so as to make for the base the side next to the base in order. This will give for each formula a group of three formulas. LIBRARY OF CONGRESS 029 944 930 LIBRARY OF CONGRESS 029 944 930