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: 74

Answer

$s=\dfrac{11\pi}{6}$

Work Step by Step

Use the inverse cosine function of a calculator in radian mode to obtain: $s =\cos^{-1}{(\frac{\sqrt3}{2})}=\frac{\pi}{6}$ The angle must be within the interval $[\frac{3\pi}{2}, 2\pi]$ Note $\cos{(\frac{\pi}{6})}=\cos{(-\frac{\pi}{6})}$. Since $\cos{\theta} = \cos{(\theta+2\pi)}$, then $\cos{(-\frac{\pi}{6})}=\cos{(-\frac{\pi}{6}+2\pi)}=\cos{(\frac{11\pi}{6})}$ Thus, $s=\dfrac{11\pi}{6}$
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