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: 67


a) $\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}$ b) $\sec^{-1}2=\frac{\pi}{3}$

Work Step by Step

a) let $\cot^{-1}(-\sqrt3)=\theta, 0\lt\theta\lt\pi$ $\cot\theta=-\sqrt3$ $\tan\theta=\frac{-1}{\sqrt3}$ $\theta=-\frac{\pi}{6}+k\pi$, where k is an integer. $\because 0\lt\theta\lt\pi,$ let $k=1, \theta=\frac{5\pi}{6}$ $\therefore \cot^{-1}(-\sqrt3)=\frac{5\pi}{6}$ b) let $\sec^{-1}2=\theta, 0\leq\theta\leq\pi, \theta\neq\frac{\pi}{2}$ $\sec\theta=2$ $\cos\theta=\frac{1}{2}$ $\theta=\pm\frac{\pi}{3}+2k\pi$, where k is an integer. $\because 0\leq\theta\leq\pi, \theta\neq\frac{\pi}{2},$ let $k=0, \theta=\frac{\pi}{3}$ $\therefore \sec^{-1}2=\frac{\pi}{3}$
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