Answer
$\frac{\sqrt 3}{2}$
Work Step by Step
Step 1. Evaluate $sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ and $cos^{-1}0=\frac{\pi}{2}$
Step 2. $sin(sin^{-1}\frac{1}{2}+cos^{-1}0)=sin(\frac{\pi}{6}+\frac{\pi}{2})=sin(\frac{2\pi}{3})=\frac{\sqrt 3}{2}$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 7 - Analytic Trigonometry - 7.5 Sum and Difference Formulas - 7.5 Assess Your Understanding - Page 488: 75
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