Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 24

Answer

$y =\displaystyle \frac{2\pi}{3}$

Work Step by Step

$y=\cos^{-1}x =\arccos x$ Domain:$[-1, 1] $ Range: $[0, \pi]$ ----------- $y$ is the number from $[0, \pi]$ such that $\displaystyle \cos y=-\frac{\sqrt{3}}{2}.$ Taking $\displaystyle \frac{\pi}{3}$ as a reference angle,$ \displaystyle \cos\frac{\pi}{3}=\frac{1}{2}$, we have (in quadrant II) $\displaystyle \cos(\frac{2\pi}{3})=-\frac{1}{2}\qquad$and$\displaystyle \quad \frac{2\pi}{3}\in [0, \pi]$ , so $y =\displaystyle \frac{2\pi}{3}$
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