Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.5 - Inverse Trigonometric Functions and Their Graphs - 5.5 Exercises - Page 445: 36


$\sin ^{-1}\left( \sin \left( \dfrac {5\pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( \dfrac {\pi }{6}\right) \right) =\dfrac {\pi }{6}$

Work Step by Step

$\sin ^{-1}\left( \sin x\right) =x\left( -\dfrac {\pi }{2}\leq x\leq \dfrac {\pi }{2}\right) $ $\dfrac {5\pi }{6} > \dfrac {\pi }{2}$ so we need to convert it $ \sin \dfrac {5\pi }{6}=\sin \dfrac {\pi }{6}$ $\Rightarrow \sin ^{-1}\left( \sin \left( \dfrac {5\pi }{6}\right) \right) =\sin ^{-1}\left( \sin \left( \dfrac {\pi }{6}\right) \right) =\dfrac {\pi }{6}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.