Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 40



Work Step by Step

Suppose $\theta =\tan (-\dfrac{\pi}{3})$ This gives: $\tan^{-1} \theta=\tan^{-1}[\tan(-\dfrac{\pi}{3})]$ The inverse property states that $\tan^{-1}(\tan a)=a $ satisfies for every $ a $ in the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ . Therefore, we have $\tan^{-1}[\tan(-\dfrac{\pi}{3})]=-\dfrac{\pi}{3}$
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