Precalculus (6th Edition) Blitzer

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

Chapter 4 - Test - Page 647: 19


$ \frac{\sqrt {9-x^2}}{3}$

Work Step by Step

Step 1. Let $u=cos^{-1}(\frac{x}{3})$; we have $cos(u)=\frac{x}{3}$ Step 2. As $x\gt0$ and $\frac{x}{3}$ is in the domain, we know $u$ is in quadrant I and $sin(u)\gt0$ Step 3. We have $sin(u)=\sqrt {1-cos^2(u)}=\sqrt {1-\frac{x^2}{9}}=\frac{\sqrt {9-x^2}}{3}$ Step 4. Thus we have $sin(cos^{-1}(\frac{x}{3}))= \frac{\sqrt {9-x^2}}{3}$
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