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: 81

Answer

$$ - \frac{7}{{25}}$$

Work Step by Step

$$\eqalign{ & \cos \left( {2\arctan \frac{4}{3}} \right) \cr & {\text{Let }}\theta = \arctan \frac{4}{3}{\text{, thus}} \cr & \tan \theta = \frac{4}{3} \cr & {\text{Recall that }}\tan \theta = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr & {\text{opposite side }} = 4 \cr & {\text{adjacent side}} = 3 \cr & {\text{hypotenuse}} = 5 \cr & \cr & {\text{We have that }}\cos \left( {2\arctan \frac{4}{3}} \right) = \cos \left( {2\theta } \right) \cr & {\text{Use the identity }}\cos 2\theta = 1 - 2{\sin ^2}\theta \cr & \cos \left( {2\arctan \frac{4}{3}} \right) = 1 - 2{\left( {\frac{4}{5}} \right)^2} \cr & \cos \left( {2\arctan \frac{4}{3}} \right) = - \frac{7}{{25}} \cr} $$
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