Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 10


$\dfrac{\sqrt 3}{2}$

Work Step by Step

The range of $\cos^{-1}{x}$ is $[0, \pi]$. Here, $\cos^{-1}{(\dfrac{1}{2})}=\dfrac{\pi}{3}$, because $\cos{(\dfrac{\pi}{3})}=\dfrac{1}{2}$ and $\dfrac{\pi}{3}$ lies in the range of $\cos^{-1}{x}$ . Therefore, we have: $\sin [\cos^{-1}{(\dfrac{1}{2})}]=\sin ({\dfrac{\pi}{3}})=\dfrac{\sqrt 3}{2}$
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