Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 34



Work Step by Step

Suppose $\theta =\cos (\dfrac{2\pi}{3})$ This gives: $\cos^{-1} \theta=\cos^{-1}[\cos (\dfrac{2\pi}{3})]$ The inverse property states that $\cos^{-1}(\cos a)=a $ satisfies for every $ a $ in the range $[0,\pi]$ . Therefore, we have $ \cos^{-1}[\cos (\dfrac{2\pi}{3})]=\dfrac{2\pi}{3}$
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