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 481: 76



Work Step by Step

Let us substitute $g^{-1}(f (\dfrac{- 5\pi}{6}))=\cos^{-1} [\sin (\dfrac{-5 \pi}{6})]$ From the unit circle, we know that: $\sin (\dfrac{- 5\pi}{6})=\dfrac{-1}{2}$ We know that the trigonometric function $\sin^{-1} (1)$ has the domain $[-1,1]$ and range $[-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$. Therefore, $\cos^{-1} [\sin (\dfrac{-5 \pi}{6})]=\dfrac{2\pi}{3}$; because $\sin \dfrac{- 5 \pi}{6}=\dfrac{-1}{2}$ and $\dfrac{2\pi}{3}$ lies in the range of $\sin^{-1} x$.
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