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: 8



Work Step by Step

Suppose $\theta = \cos^{-1} (\dfrac{\sqrt 2}{2})$ This gives: $\cos \theta= \dfrac{\sqrt 2}{2}$ Now, $\theta = \cos^{-1} (\dfrac{\sqrt 2}{2})=\dfrac{\pi}{4}$ Therefore, $\cos^{-1} (\dfrac{\sqrt 2}{2})=\dfrac{\pi}{4}$
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