Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 709: 88

Answer

$$ - \frac{{16}}{{65}}$$

Work Step by Step

$$\eqalign{ & \cos \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cos }^{ - 1}}\frac{5}{{13}}} \right) \cr & {\text{Let }}\theta = {\sin ^{ - 1}}\frac{3}{5}{\text{ and }}\beta = {\cos ^{ - 1}}\frac{5}{{13}} \cr & \cr & {\text{Therefore,}} \cr & \theta = {\sin ^{ - 1}}\frac{3}{5} \to \sin \theta = \frac{3}{5} = \frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} \cr & {\text{adjacent side}} = \sqrt {{5^2} - {3^2}} = 4 \cr & and \cr & \beta = {\cos ^{ - 1}}\frac{5}{{13}} \to \cos \theta = \frac{5}{{13}} = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr & {\text{opposite side}} = \sqrt {{{13}^2} - {5^2}} = 12 \cr & \cr & \cos \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cos }^{ - 1}}\frac{5}{{13}}} \right) = \cos \left( {\theta + \beta } \right) \cr & {\text{Use the cosine of a sum}} \cr & = \cos \theta \cos \beta - \sin \theta \sin \beta \cr & = \left( {\frac{4}{5}} \right)\left( {\frac{5}{{13}}} \right) - \left( {\frac{3}{5}} \right)\left( {\frac{{12}}{{13}}} \right) \cr & {\text{Simplify}} \cr & = - \frac{{16}}{{65}} \cr} $$
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