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: 66


a) $\sin^{-1}(-\frac{1}{\sqrt2})=-\frac{\pi}{4}$ b) $\cos^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6}$

Work Step by Step

a) let $\sin^{-1}(-\frac{1}{\sqrt2})=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ $\sin\theta=-\frac{1}{\sqrt2}$ $\theta=-\frac{\pi}{4}+2k\pi, -\frac{3\pi}{4}+2k\pi$, where k is an integer $\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2},$ let $k=0, \theta=-\frac{\pi}{4}$ $\therefore \sin^{-1}(-\frac{1}{\sqrt2})=-\frac{\pi}{4}$ b) let $\cos^{-1}\frac{\sqrt3}{2}=\theta, 0\leq\theta\leq\pi$ $\cos\theta=\frac{\sqrt3}{2}$ $\theta=\frac{\pi}{6}+2k\pi, -\frac{\pi}{6}+2k\pi$, where k is an integer $\because 0\leq\theta\leq\pi, $ let $k=0, \theta=\frac{\pi}{6}$ $\therefore \cos^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6}$
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