Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Cumulative Review Exercises - Page 801: 11


$\frac{\sqrt 5}{5}$

Work Step by Step

Step 1. Let $u=tan^{-1}\frac{1}{2}, 0\lt u\lt \frac{\pi}{2}$; we have $tan(u)=\frac{1}{2}$ Step 2. Construct a right triangle with acute angle $u$. Opposite side $=1$ and adjacent side $=2$. We have the hypotenuse $=\sqrt {1^2+2^2}=\sqrt 5$. Thus $tan(u)=\frac{1}{2}$ and $sin(u)=\frac{\sqrt 5}{5}$ Step 3. Thus we have $sin(tan^{-1}\frac{1}{2})=sin(u)=\frac{\sqrt 5}{5}$
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