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: 64


a) $\tan^{-1}\sqrt3=\frac{\pi}{3}$ b) $\arctan(-1)=-\frac{\pi}{4}$

Work Step by Step

a) let $\tan^{-1}\sqrt3=\theta , -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$ $\tan\theta=\sqrt3$ $\theta=\frac{\pi}{3}+k\pi$ $\because -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, let $k=0, \theta=\frac{\pi}{3}$ $\therefore \tan^{-1}\sqrt3=\frac{\pi}{3}$ b) let $\arctan(-1) =\theta, -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$ $\tan\theta=-1$ $\theta=-\frac{\pi}{4}+k\pi$ $\because -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, let $k=0, \theta=-\frac{\pi}{4}$ $\therefore \arctan(-1)=-\frac{\pi}{4}$
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