Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Chapter Test - Page 507: 19

Answer

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

Work Step by Step

Letting $cos^{-1}\frac{3}{5}=t, 0\lt t\lt\frac{\pi}{2}$, we have $cos(t)=\frac{3}{5}$ Thus $sin(\frac{t}{2})=\sqrt {\frac{1-cos(t)}{2}}=\sqrt {\frac{1-\frac{3}{5}}{2}}=\frac{\sqrt 5}{5}$
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