734 CHAPTER 7 Trigonometric Identities and Equations (b) Let B = arcsin 2 5 , so that sin B = 2 5 . Sketch angle B in quadrant I. Use the Pythagorean theorem to find the unknown side of the triangle. See Figure 26. tan a2 arcsin 2 5b Given expression = 2 A 21 21B 1 - A 21 21B 2 = 41 21 1 - 4 21 Multiply and apply the exponent. = 41 21 # 121 121 17 21 = 4121 21 17 21 Multiply in the numerator. = 4221 17 Divide; a b c d = a b , c d = a b # d c S Now Try Exercises 79 and 87. y x 0 2 5 B √21 Figure 26 Use tan 2B = 2 tan B 1 - tan2 B with tan B = 21 21 from Figure 26. Rationalize in the numerator. Subtract in the denominator. y x 0 u, u > 0 u, u < 0 1 U U √u2 + 1 √u2 + 1 Figure 27 EXAMPLE 7 Writing Function Values inTerms of u Write each trigonometric expression as an algebraic expression in u. (a) sin1tan-1 u2 (b) cos12 sin-1 u2 SOLUTION (a) Let u = tan-1 u, so tan u = u. Because - p 2 6tan-1 u 6p 2 , sketch u in quadrants I and IV and label two triangles, as shown in Figure 27. Sine is given by the quotient of the side opposite and the hypotenuse, so we have sin1tan-1 u2 = sin u = u 2 u2 + 1 = u 2 u2 + 1 # 2u2 + 1 2u2 + 1 = u2u2 + 1 u2 + 1 . The result is positive when u is positive and negative when u is negative. Rationalize the denominator. (b) Let u = sin-1 u, so that sin u = u. cos12 sin-1 u2 = cos 2u = 1 - 2 sin2 u = 1 - 2u2 Double-angle identity S Now Try Exercises 95 and 99. To support our algebraic work, we could enter cos Aarctan23 + arcsin 1 3B from Example 6(a) into a calculator. The approximation 0.1827293862 is the same approximation we obtain when we enter the exact value 2 22 - 23 6 . Similarly, we obtain the same approximation when we evaluate tan A2 arcsin 2 5B and 4 221 17 , supporting our answer in Example 6(b).
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