Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 420: 2

Answer

$a.\displaystyle \quad -\frac{\pi}{4}$ $b.\displaystyle \quad\frac{\pi}{3}$ $c.\displaystyle \quad -\frac{\pi}{6}$

Work Step by Step

$y=\tan^{-1}x$ is the number in $(-\pi/2, \pi/2)$ for which $\tan y=x.$ Use reference angles in the 1st quadrant, keeping in mind that tan is an odd function. $\left\{\begin{array}{llll} \tan\frac{\pi}{4}=1 & \Rightarrow & \tan(-\frac{\pi}{4})=-1 & \Rightarrow\tan^{-1}(-1)=-\frac{\pi}{4}\\ & & & \\ \tan\frac{\pi}{3}=\sqrt{3} & \Rightarrow & & \Rightarrow\tan^{-1}(\sqrt{3})=\dfrac{\pi}{3}\\ & & & \\ \tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} & \Rightarrow & \tan(-\frac{\pi}{6})=-\dfrac{1}{\sqrt{3}} & \Rightarrow\tan^{-1}(-\frac{1}{\sqrt{3}})=-\dfrac{\pi}{6} \end{array}\right.$
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