Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 589: 71



Work Step by Step

Use the inverse tangent function of a calculator in radian mode to obtain: $s=\tan^{-1}{(\sqrt3)} = \frac{\pi}{3}$ The angle must be in the given interval, which is actually in Quadrant III. Note that for an angle $\theta$ in Quadrant I, then $\tan{\theta} = \tan{(\pi+\theta)}$ Thus, $\tan{\frac{\pi}{3}} = \tan{(\pi+\frac{\pi}{3})}=\tan{\frac{4\pi}{3}}$ Therefore, then angle must be $\color{blue}{\dfrac{4\pi}{3}}$.
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