## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Test - Page 647: 19

#### Answer

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

#### 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 $sin(cos^{-1}(\frac{x}{3}))= \frac{\sqrt {9-x^2}}{3}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.