754 CHAPTER 7 Trigonometric Identities and Equations Because the function y = 3 cos 2x is periodic, with period p, there are infinitely many domain values (x-values) that will result in a given range value (y-value). For example, the x-values 0 and p both correspond to the y-value 3. See Figure 38. The restriction 0 … x … p 2 given in the original problem ensures that this function is one-to-one and, correspondingly, that x = 1 2 arccos y 3 has a one-to-one relationship. Thus, each y-value in 3-3, 34 substituted into this equation will lead to a single x-value. S Now Try Exercise 9. y 0 3 (0, 3) (P, 3) –3 p p 4 p 2 3p 4 x ( , –3) P 2 0 ≤ x ≤ y = 3 cos 2x P 2 Figure 38 Solution of Inverse Trigonometric Equations EXAMPLE 2 Solving an Equation Involving an Inverse Trigonometric Function Solve 2 arcsin x = p. SOLUTION Solve first for arcsin x and then for x. 2 arcsin x = p Original equation arcsin x = p 2 Divide by 2. x = sin p 2 Definition of arcsine x = 1 arcsin 1 = p 2 CHECK 2 arcsin x = p Original equation 2 arcsin 1≟p Let x = 1. 2 a p 2b ≟p Substitute the inverse value. p = p ✓ True The solution set is 516. S Now Try Exercise 25. EXAMPLE 3 Solving an Equation Involving Inverse Trigonometric Functions Solve cos-1 x = sin-1 1 2 . SOLUTION Let sin-1 1 2 = u. Then sin u = 1 2 , and for u in quadrant I we have cos-1 x = sin-1 1 2 Original equation cos-1 x = u Substitute. cos u = x. Alternative form Sketch a triangle and label it using the facts that u is in quadrant I and sin u = 1 2 . See Figure 39. Because x = cos u, we have x = 23 2 . The solution set is U 23 2 V . S Now Try Exercise 31. 0 2 1 u x y √3 Figure 39
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