Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises: 77

Answer

$y= -\displaystyle \frac{\pi}{3}$

Work Step by Step

By definition of arctan $(\tan^{-1})$, y is the number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ such that $\tan y=-\sqrt{3}.$ In quadrant I, $\displaystyle \tan(\frac{\pi}{3})=\sqrt{3}$, so in quadrant IV, $\displaystyle \tan(-\frac{\pi}{3})=-\sqrt{3}.$ $-\displaystyle \frac{\pi}{3}\in (-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ so $y= -\displaystyle \frac{\pi}{3}$
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