Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 68: 65


a) $\csc^{-1}\sqrt2=\frac{\pi}{4}$ b) $\arcsin1=\frac{\pi}{2}$

Work Step by Step

a) let $\csc^{-1}\sqrt2=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}, \theta\neq0$ $\csc\theta=\sqrt2$ $\sin\theta=\frac{1}{\sqrt2}$ $\theta=\frac{\pi}{4}+2k\pi, \frac{3\pi}{4}+2k\pi$ where k is an integer $\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}, \theta\neq0$, let $k=0, \theta=\frac{\pi}{4}$ $\therefore \csc^{-1}\sqrt2=\frac{\pi}{4}$ b) let $\arcsin1=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ $\sin\theta=1$ $\theta=\frac{\pi}{2}+2k\pi$ where k is an integer $\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, let $k=0, \theta=\frac{\pi}{2}$ $\therefore \arcsin1=\frac{\pi}{2}$
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