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: 4

Answer

$a.\displaystyle \quad\frac{5\pi}{6}$ $b.\displaystyle \quad\frac{\pi}{2}$

Work Step by Step

$ a.\quad$ $y=\cot^{-1}x,\ (x\in \mathbb{R}) \ \Leftrightarrow\ \cot y=x\ $ and $\ y\in(0, \pi)$ $y=\displaystyle \frac{\pi}{6}\in(0, \pi)$ is such that $\displaystyle \cot\frac{\pi}{6}=+\sqrt{3}$, By symmetry, $\displaystyle \cot\frac{5\pi}{6}=-\sqrt{3} ,$ which means that $\quad\cot^{-1}(-\sqrt{3})$=$ \displaystyle \frac{5\pi}{6}$ $ b.\quad$ $\arcsin 1=\sin^{-1}1,$ $\sin^{-1}x=y \ \Leftrightarrow\ \sin y=x\ $ and $\ y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}].$ $y=\displaystyle \frac{\pi}{2}\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$ is such that $\displaystyle \sin\frac{\pi}{2}=1$, so $\displaystyle \arcsin 1=\frac{\pi}{2}$
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