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: 41


$-\dfrac{ \pi}{8}$

Work Step by Step

Value of certain composite functions formula: $f^{-1}(f(x)= \sin^{-1}{(\sin{x})} = x \hspace{25pt} -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$ Note that $\dfrac{9 \pi}{8} \text{ isn't in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$. Recall that: $\sin{\theta} = \sin{(\pi - \theta)}$ Thus, $\sin{\left(\dfrac{9 \pi}{8} \right)} = \sin{\left(\pi - \dfrac{9 \pi}{8} \right) }= \sin{\left(-\dfrac{\pi}{8} \right)}$ Since $-\dfrac{\pi}{8} \text{ is in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$, then $\sin^{-1} {\left[\sin{\left(-\dfrac{ \pi}{8} \right)} \right]} = -\dfrac{ \pi}{8}$ Therefore, $\sin^{-1} {\left[\sin{\left(\dfrac{ 9\pi}{8} \right)} \right]} = \boxed{-\dfrac{ \pi}{8}}$
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