Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 53


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

Work Step by Step

Let $u=tan^{-1}~2$. Then: $tan~u=2$ The range $tan^{-1}x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $tan^{-1}~2\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant) $sec^2u=1+tan^2u$ $sec^2u=1+2^2=5$ $sec~u=\sqrt 5~~$ (First Quadrant) $cos(tan^{-1}~2)=cos~u=\frac{1}{sec~u}=\frac{1}{\sqrt 5}=\frac{1}{\sqrt 5}\frac{\sqrt 5}{\sqrt 5}=\frac{\sqrt 5}{5}$
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