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492 Chapter 18 ◆ Trigonometric Identities and Equations 19. sin cos(30 ) sin(60 ) 20. cos(x y) cos(x y) 2 cos x cos y 21. cos x sin p x q p sin x q 6 6 22. tan(360 ) tan 23. cos(60 ) sin(330 ) 0 sin 4x cos 4x 24. sin 5x sec x csc x sin x cos x sin(x y) tan x tan y 25. sin(x y) tan x tan y 26. cos(x 60) cos(60 x) cos x cos(x y) 27. tan y cot x sin x cos y 28. cotp x q tanp x q 0 4 4 cos sin 29. tan( 45) cos sin cot cot 1 30. cot( ) cot cot 1 tan x 31. tan p x q 1 tan x 4 Express as a single sine function. 32. y 47.2 sin t 64.9 cos t 33. y 8470 sin t 7360 cos t 34. y 1.83 sin t 2.74 cos t 35. y 84.2 sin t 74.2 cos t 18–3 Functions of Double Angles Sine of 2 An equation for the sine of 2 is easily derived by setting in Eq. 167. sin( ) sin cos cos sin which is rewritten in the following form: Sine of Twice an Angle sin 2 2 sin cos 170 Cosine of 2 Similarly, setting in Eq. 168, we have cos( ) cos cos sin sin which is also given as follows: Cosine of Twice an Angle cos 2 cos2 sin2 171a Section 18–3 ◆ 493 Functions of Double Angles There are two alternative forms for Eq. 171a. From Eq. 164, cos2 1 sin2 Substituting into Eq. 171a yields cos 2 1 sin2 sin2 cos 2 1 sin2 Cosine of Twice an Angle 171b We can similarly use Eq. 164 to eliminate the sin2 term from Eq. 171a. cos 2 cos2 (1 cos2 ) Thus: cos 2 2 cos2 1 Cosine of Twice an Angle ◆◆◆ 171c Example 17: Express sin 3x in terms of the single angle x. Solution: We can consider 3x to be x 2x, and, using Eq. 167, we have sin 3x sin(x 2x) sin x cos 2x cos x sin 2x We now replace cos 2x by cos2 x sin2 x, and sin 2x by 2 sin x cos x. sin 3x sin x(cos2 x sin2 x) cos x(2 sin x cos x) sin x cos2 x sin3 x 2 sin x cos2 x 3 sin x cos2 x sin3 x ◆◆◆ ◆◆◆ Example 18: Prove that cos 2A sin(A B) 0 where A and B are the two acute angles of a right triangle. Solution: By Eqs. 171a and 167, cos 2A sin(A B) cos2 A sin2 A sin A cos B cos A sin B But, by the cofunctions of Eq. 154, cos B sin A and sin B cos A So cos 2A sin(A B) cos2 A sin2 A sin A sin A cos A cos A cos2 A sin2 A sin2 A cos2 A 0 ◆◆◆ ◆◆◆ Example 19: Simplify the expression sin 2x 1 cos 2x Solution: By Eqs. 170 and 171c, sin 2x 1 cos 2x 2 sin x cos x 1 2 cos2 x 1 2 sin x cos x 2 cos2 x sin x tan x cos x ◆◆◆ 494 Chapter 18 ◆ Trigonometric Identities and Equations Tangent of 2␣ Setting in Eq. 169 gives tan tan tan ( ) 1 tan tan Therefore: Tangent of Twice an Angle ◆◆◆ 2 tan x tan 2 = 1 tan2 x 172 Example 20: Prove that 2 cot x csc2 x 2 tan 2x Solution: By Eq. 172, 2 tan tan 2x 1 tan2 and, by Eq. 152c, 2 cot x tan 2x 1 1 cot2 x Multiply numerator and denominator by cot2 x. 2 cot x tan 2x cot2 x 1 Then, by Eq. 166, 2 cot x 2 cot x tan 2x 2 (csc x 1) 1 csc2 x 2 Common Error Exercise 3 ◆ The sine of twice an angle is not twice the sine of the angle. Nor is the cosine (or tangent) of twice an angle equal to twice the cosine (or tangent) of that angle. Remember to use the formulas from this section for all of the trig functions of double angles. Functions of Double Angles Simplify. 1. 2 sin2 x cos 2x 2. 2 sin 2 cos 2 2 tan x 3. 1 tan2 x 2 sec2 x 4. sec2 x sin 6x cos 6x 5. sin 2x cos 2x ◆◆◆ Section 18–4 ◆ 495 Functions of Half-Angles Prove. 2 tan 6. tan 2 1 tan2 1 tan2x 7. cos 2x 1 tan2x 8. 2 csc 2 tan cot 9. 2 cot 2x cot x tan x 1 cot2 x 10. sec 2x cot2 x 1 1 tan x cos 2x 11. tan x 1 sin 2x 1 sin 2 sin 12. tan 1 cos cos 2 2 cos 2x 13. 1 cot x sin 2x 2 sin2 x 14. 4 cos3 x 3 cos x cos 3x sin 2 1 15. sin cos cos sin cot2 x 1 16. cot 2x 2 cot x If A and B are the two acute angles in a right triangle, show that: 17. sin 2A sin 2B. 18. tan(A B) cot 2A. 19. sin 2A cos(A B). 18–4 Functions of Half-Angles Sine of ␣冒2 The double-angle formulas derived in Sec. 18–3 can also be regarded as half-angle formulas, because if one angle is double another, the second angle must be half the first. Starting with Eq. 171b, we obtain cos 2 1 2 sin2 We solve for sin . 2 sin2 1 cos 2 sin 兹 1 cos 2 2 For emphasis, we replace by 冒2. Sine of Half an Angle sin 2 兹 1 cos 2 173 The sign in Eq. 173 (and in Eqs. 174 and 175c as well) is to be read as plus or minus, but not both. This sign is different from the sign in the quadratic formula, for example, where we took both the positive and the negative values. The reason for this difference is clear from Fig. 18–4, which shows a graph of sin 冒2 and a graph of 兹 (1 cos )冒2 . Note that the two curves are the same only when sin 冒2 is positive. When sin 冒2 is negative, it is necessary to use the negative of 兹 (1 cos )冒2 . This occurs when 冒2 is in the third or fourth quadrant. Thus we choose the plus or the minus according to the quadrant in which 冒2 is located.