Resolving a portion of the expression into factors, and applying the formula sin2 A+ cos2 A = 1, sin60+ cos 0+3 sin2 0 cos2 0. = (sin20+ cos2 ) (sin1 6-sin2 0 cos2 + cos1 0) + 3 sin2 0 cos2 0 sin40-sin20 cos2 0 + cos 0+3 sin2 0 cos2 0 6. Express sin A in terms of tan A, and cos A in terms of cosec A. 7. Express tan A in terms of cos A, and cot A in terms of sec A. 8. Express sec A in terms of sin A, and cosec A in terms of versin A. 9. Given tan A=3, find all the other trigonometrical ratios of A. 10. Given sin A=5, find tan A, cos A, versin A. 11. Given sec A = 2/2, find sin A, cot A. Prove the formulæ in 12-31. 12. 1+2 sin cos 0 = (sin + cos 0)2. sin2 A-sin2 B. 25. cos2 A cos2 B-sin2 A sin2 B = 26. 2 (sin60+ cos ◊) — 3 (sin1 + cos1 0) + 1 = 0. 27. sin2 0 tan + cos2 0 cot 0+2 sin cos 0: = tan 0+ cot 0. 31. cot 0 (1-cos 0) + tan e = cosec 0 (sec 0 − 1) + sin 0. 32. If tan + sin 0 = m, and tan 0—sin 0 = n; then m2 —n2 = 4√mn. 33. Show that if C be the chord subtending an angle 2a at the centre of a circle of radius r, then C = 2r sin a. CHAPTER III. COMPLEMENTS.-NUMERICAL VALUES OF CERTAIN RATIOS. 17. Def.-If the sum of two angles be a right angle, each of them is called the complement of the other. The same thing is expressed by saying that A and 90°-A are complementary to each other; or that a and ẞ are complementary when a +6: = Π 2 It follows that, in any right-angled triangle ABC, in which is the right angle, A and B are complements ; for by Euc. I. 32, A+B=90°. 18. We shall now obtain additional formulæ worthy of careful attention; numbering some which should be committed to memory. From these the others are easily deduced whenever required. Let A and B be the acute angles of any right-angled triangle ABC; then .. sin 4 = cos B, or sin 4 = cos (90°-4). cos Asin (90°-A). S The same mode of proof will also show that (1). tan A cot (90°-4), and cot A = tan (90°-A)...(2), sec A= =cosec (90°— A), and cosec A=sec (90°— A)...(3). Or we might have proceeded in these last cases after the following manner : The designations cosine, cotangent, cosecant, have their origin in the facts just proved. 19. To find the numerical values of the trigonometrical ratios of the angle 45°. Since when A is any angle, sin? A+cos2 A = 1, 20. To find the values of the trigonometrical ratios of 30° and 60°. Let the triangle ABC be equilateral, and let AB be drawn bisecting the angle BAC; then by Euc. I., Props. 4, 32, it is easily shown that .. sin 30° = B Hence and cos 30° = √/1—sin230° = √3; tan 30° = The values of cot 30°, Since cos 30° √3* sec 30°, cosec 30° are obvious. sin A = cos (90°—A), we have from the above 21. The following are instances of Exercises on the foregoing formulæ : (2.) Find A, when sin A = cos 24. cos 24 = sin (90°—24), .. sin A = sin (90°—2A); .. A = = 90°—24, and 34 (3.) Show that .. A = 30°. = 90°; cos 45° + cos 30° sec 45°+cot 30° = sec 45°. -cot 45°° 1 √2 2 = √2+√3 sec 45° + cot 30° (4.) Solve the equation sec Obs. If there be more functions than one involved in the equation, transform until there is one only; that function is the unknown quantity of the equation. The remaining part of the process is obvious from the Examples. |