733 7.5 Inverse Circular Functions (b) Let A = cos-1A - 5 13B . Then cos A = - 5 13 . Because cos-1 x for a negative value of x is in quadrant II, sketch A in quadrant II. See Figure 24. tan acos-1 a- 5 13bb = tan A = - 12 5 S Now Try Exercises 75 and 77. Figure 24 y x 0 13 –5 12 A A = cos–1(– ) 5 13 Figure 23 y x 0 3 2 U U = tan –1 2 3 √13 EXAMPLE 6 Finding Function Values Using Identities Evaluate each expression without using a calculator. (a) cos aarctan 23 + arcsin 1 3b (b) tan a2 arcsin 2 5b SOLUTION (a) Let A=arctan 23 and B=arcsin 1 3 . Therefore, tan A= 23 and sin B= 1 3 . Sketch both A and B in quadrant I, as shown in Figure 25, and use the Pythagorean theorem to find the unknown side of each triangle. Figure 25 y x 0 2 √3 1 A y x 0 1 3 B 2√2 cos aarctan23 + arcsin 1 3b Given expression = cos1A + B2 Let A = arctan23 and B = arcsin 1 3 . = cos A cos B - sin A sin B Cosine sum identity = 1 2 # 2 22 3 - 23 2 # 1 3 Substitute values using Figure 25. = 222 - 23 6 Multiply and write as a single fraction.
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