Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.6 Inverse Trigonometric Functions - 6.6 Exercises - Page 481: 6

Answer

$a.\displaystyle \quad-\frac{\pi}{4}$ $b.\displaystyle \quad\frac{\sqrt{3}}{2}$

Work Step by Step

$ a.\quad$ $\displaystyle \tan\frac{3\pi}{4}=-1$ $\tan^{-1}(-1)=y \Leftrightarrow \tan y=-1$ and $y\in(- \displaystyle \frac{\pi}{2},\frac{\pi}{2})$ In $(- \displaystyle \frac{\pi}{2},\frac{\pi}{2}),\ y=-\displaystyle \frac{\pi}{4}$ is such that $\displaystyle \tan(-\frac{\pi}{4})=-1$, so $\displaystyle \tan^{-1}\tan(\frac{3\pi}{4})=\tan^{-1}(-1)=-\frac{\pi}{4}$ $ b.\quad$ $\arcsin x=\sin^{-1}0.5=y \ \Leftrightarrow\ \sin y=0.5\ $ and $\ y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$ In $[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}],\ y=\displaystyle \frac{\pi}{6}$ is such that $\sin y=0.5$, so $\displaystyle \cos(\arcsin\frac{1}{2})=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$
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