Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

Chapter 5 - Section 5.5 - Inverse Trigonometric Functions and Their Graphs - 5.5 Exercises: 38

Answer

$\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {7 \pi }{6}\right) \right) =-\dfrac {\pi }{6}$

Work Step by Step

$\sin ^{-1}\left( \sin x\right) =x\left( -\dfrac {\pi }{2}\leq x\leq \dfrac {\pi }{2}\right)$ $\dfrac {7\pi }{6} > \dfrac {\pi }{2}$ so we need to convert it $\sin \dfrac {7\pi }{6}=\sin (-\dfrac {\pi }{6})$. ($\sin \alpha =\sin \left( \pi -\alpha \right)$) $\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {7 \pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( -\dfrac {\pi }{6}\right) \right) =-\dfrac {\pi }{6}$

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