704 CHAPTER 7 Trigonometric Identities and Equations In the same way, sin B = - 12 13 . Now find sin1A + B2. sin1A + B2 = sin A cos B + cos A sin B Sine sum identity = 4 5 a- 5 13b + a- 3 5b a- 12 13b = - 20 65 + 36 65 Multiply. sin1A + B2 = 16 65 Add. (b) To find tan1A + B2, use the values of sine and cosine from part (a), sin A = 4 5 , cos A = - 3 5 , sin B = - 12 13 , and cos B = - 5 13 , to obtain tan A and tan B. tan A = sin A cos A = 4 5 - 3 5 = 4 5 , a- 3 5b = 4 5 # a- 5 3b tan A = - 4 3 Substitute the given values for sin A and cos B and the values found for cos A and sin B. tan1A + B2 = tan A + tan B 1 - tan A tan B Tangent sum identity = - 4 3 + 12 5 1 - A - 4 3B A 12 5 B Substitute. = 16 15 1 + 48 15 = 16 15 63 15 1 = 15 15; Add in the denominator. = 16 15 , 63 15 Simplify the complex fraction. = 16 15 # 15 63 Definition of division tan1A + B2 = 16 63 Multiply. (c) sin1A + B2 = 16 65 and tan1A + B2 = 16 63 See parts (a) and (b). Both are positive. Therefore, A + B must be in quadrant I, because it is the only quadrant in which both sine and tangent are positive. S Now Try Exercise 83. tan B = sin B cos B = - 12 13 - 5 13 = - 12 13 , a- 5 13b = - 12 13 # a- 13 5 b tan B = 12 5 Perform the indicated operations; -4 3 + 12 5 = - 20 15 + 36 15 = 16 15
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