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: 43

Answer

$\frac{1}{\sqrt {3}}$

Work Step by Step

Let $\sin^{-1}(\frac{1}{2})=\theta$ Then, $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$ $⇒\theta=\frac{\pi}{6}$. Therefore, $\tan(\sin^{-1}\frac{1}{2})= \tan(\frac{\pi}{6})= \frac{1}{\sqrt 3}$
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