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 473: 19



Work Step by Step

$y = \sin^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \sin{y}$ $\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$ Let $\theta = \sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$ Since $\theta = \sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)}$, then $\sin{\theta} = \dfrac{\sqrt{2}}{2}$ Referring to Table 1 on page 466: The only angle in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$ whose sine is $\dfrac{\sqrt{2}}{2}$ is $\dfrac{\pi}{4}$. Therefore, $\sin^{-1} {\left(\dfrac{\sqrt{2}}{2} \right)} = \dfrac{\pi}{4}$.
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