Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 88

Answer

$cos~(2~tan^{-1}~(-2)) = -\frac{3}{5}$

Work Step by Step

Let $~~\theta = tan^{-1}(-2)$ Then: $~~tan~\theta = -2 = \frac{-2}{1}$ Then: $~~sin~\theta = \frac{-2}{\sqrt{(-2)^2+(1)^2}} = -\frac{2}{\sqrt{5}}$ We need to find $cos~2\theta$: $cos~2\theta = 1-2~sin^2~\theta$ $cos~2\theta = 1-2~(-\frac{2}{\sqrt{5}})^2$ $cos~2\theta = 1-\frac{8}{5}$ $cos~2\theta = -\frac{3}{5}$ Therefore, $~~cos~(2~tan^{-1}~(-2)) = -\frac{3}{5}$
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