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: 75

Answer

$\dfrac{\pi}{6}$

Work Step by Step

We know that $ f(x)=\sin(x), g(x)=\cos(x)$ Thus we can write: $g^{-1}(f (\dfrac{-4\pi}{3}))=\cos^{-1} (\sin (\dfrac{-4 \pi}{3}))$ From the unit circle, we know that: $\sin (\dfrac{-4\pi}{3})=\dfrac{-1}{2}$ Therefore, $\cos^{-1} (\sin (\dfrac{-4 \pi}{3}))=\cos^{-1}(\dfrac{-1}{2})=\dfrac{\pi}{6}$ Note that we restrict the range of $\cos^{-1}$ to $[0,\pi]$.
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