Answer
$\frac{\sqrt 5}{5}$.
Work Step by Step
Given $tan\theta=\frac{1}{2}, -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, let $x=2, y=1$, we have $r=\sqrt {2^2+1^2}=\sqrt 5$ and $sin\theta=\frac{y}{r}=\frac{\sqrt 5}{5}$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 3
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