755 7.7 Equations Involving Inverse Trigonometric Functions x y 1 u x 0 √1 – x2 Figure 40 Square each factor. Choose the positive square root, x 70. EXAMPLE 4 Solving an InverseTrigonometric Equation Using an Identity Solve arcsin x - arccos x = p 6 . SOLUTION Isolate one inverse function on one side of the equation. arcsin x - arccos x = p 6 Original equation arcsin x = arccos x + p 6 Add arccos x. (1) x = sin aarccos x + p 6b Definition of arcsine Let u = arccos x. The arccosine function yields angles in quadrants I and II, so 0 … u … p by definition. x = sin au + p 6b Substitute. x = sin u cos p 6 + cos u sin p 6 Sine sum identity (2) Use equation (1) and the definition of the arcsine function. - p 2 … arccos x + p 6 … p 2 Range of arcsine is C - p 2 , p 2 D . - 2p 3 … arccos x … p 3 Subtract p 6 from each part. Because both 0 …arccos x …p and - 2p 3 …arccos x … p 3 , the intersection yields 0 … arccos x … p 3 . This places u in quadrant I, and we can sketch the triangle in Figure 40. From this triangle we find that sin u = 21 - x2. Now substitute into equation (2) using sin u = 21 - x2, sin p 6 = 1 2 , cos p 6 = 23 2 , and cos u = x. x = sin u cos p 6 + cos u sin p 6 (2) x = A 21 - x2 B 23 2 + x # 1 2 Substitute. 2x = A 21 - x2 B 23 + x Multiply by 2. x = A 23 B 21 - x2 Subtract x; commutative property x 2 = 311 - x22 Square each side; 1ab22 = a2 b2 x 2 = 3 - 3x2 Distributive property x 2 = 3 4 Add 3x2. Divide by 4. x = B3 4 Take the square root on each side. x = 23 2 Quotient rule: 3n a b = 2n a2 n b
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