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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. 496 Chapter 18 ◆ Trigonometric Identities and Equations y 1 y = sin α 2 0.5 0 180˚ 360˚ 540˚ 720˚ α (deg) y 0.5 0 y = + 1 − cos α 2 α (deg) FIGURE 18–4 ◆◆◆ Example 21: Given that cos 200 0.9397, find sin 100. Solution: From Eq. 173, 1 (0.9397) 200 1 cos 200 sin 0.9848 2 2 2 Since 100 is in the second quadrant, the sine is positive, so we drop the minus sign and get 兹 兹 sin 100 0.9848 ◆◆◆ Cosine of 冒2 Similarly, starting with Eq. 171c, we have cos 2 2 cos2 1 Then we solve for cos . 2 cos2 1 cos 2 cos 兹 1 cos 2 2 We next replace by 冒2 and obtain the following: Here also, we choose the plus or the minus according to the quadrant in which /2 is located. Cosine of Half an Angle cos 2 兹 1 cos 2 174 Section 18–4 ◆ 497 Functions of Half-Angles Tangent of 冒2 There are three formulas for the tangent of a half-angle; we will show the derivation of each in turn. First, from Eq. 162, sin 2 tan 2 cos 2 Multiply numerator and denominator by 2 sin(冒2). 2 sin2 2 tan 2 2 sin cos 2 2 Then, by Eqs. 171b and 170, we have the following: 1 cos tan 2 sin Tangent of Half an Angle 175a Another form of this identity is obtained by multiplying numerator and denominator by 1 cos . 1 cos 1 cos tan 2 1 cos sin 1 cos2 sin2 sin (1 cos ) sin (1 cos ) also written in the following form: Tangent of Half an Angle sin tan 2 1 cos 175b We can obtain a third formula for the tangent by dividing Eq. 173 by Eq. 174. 1 cos sin 2 2 tan 2 cos 1 cos 2 2 兹 兹 Tangent of Half an Angle ◆◆◆ tan 2 兹 1 cos 1 cos Example 22: Prove that 1 sin2 3 cos 2 3 cos 1 cos2 2 175c 498 Chapter 18 ◆ Trigonometric Identities and Equations Solution: By Eqs. 173 and 174, 1 cos 1 sin2 1 2 2 1 cos 1 cos2 1 2 2 Multiply numerator and denominator by 2. 2 1 cos 2 1 cos 3 cos 3 cos Exercise 4 ◆ ◆◆◆ Functions of Half-Angles Prove each identity. 1. 2 sin2 cos 1 2 x x 2. 4 cos2 sin2 1 cos2 x 2 2 x sec x 1 3. 2 cos2 sec x 2 4. cot csc cot 2 sin 5. sin sec cos 1 2 2 cos2 cos 2 6. 1 sin2 2 7. sec tan tan 1 2 x x 2 8. sin x 1 p cos sin q 2 2 9. 2 sin sin 2 4 sin cos2 2 1 tan2 2 11. cos 1 tan2 2 1 cos x x 10. tan sin x 2 In right triangle ABC, show the following: a A 12. tan 2 bc 18–5 A 13. sin 2 cb 兹 2c Trigonometric Equations Solving Trigonometric Equations One use for the trigonometric identities we have just studied is in the solution of trigonometric equations. A trigonometric equation will usually have an infinite number of roots. However, it is customary to list only nonnegative values less than 360 that satisfy the equation.