Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 11



Work Step by Step

RECALL: $y=\sec^{-1}{x} \longrightarrow \cos{y}=\frac{1}{x}$, $y$ is in the interval $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ Thus, $y=\sec^{-1}{(-2)}$ implies that $\cos{y}=\frac{1}{-2}=-\frac{1}{2}$. Note that $\cos{(\frac{2\pi}{3})}=-\frac{1}{2}$. Therefore, $y=\sec^{-1}{(-2)}\longrightarrow y=\frac{2\pi}{3}$
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