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 6.1 The Inverse Sine, Cosine, and Tangent Functions - 6.1 Assess Your Understanding - Page 474: 62



Work Step by Step

Divide both sides by $2$ to obtain: $\cos^{-1}{x} = \dfrac{\pi}{2}$ Recall: $y = \cos^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \cos{y}$ $\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} 0 \leq y \leq \pi$ Thus, using the definition above gives: $\cos^{--1}{x}=\dfrac{\pi}{2} \longrightarrow x = \cos{\dfrac{\pi}{2}}$ $x =\cos{\dfrac{\pi}{2}} = \boxed{0}$
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