Precalculus (10th Edition)

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

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 28

Answer

$\frac{\sqrt 7}{3}$

Work Step by Step

Step 1. Letting $sin^{-1}\frac{\sqrt 2}{3}=t$, we have $sin(t)=\frac{\sqrt 2}{3}$ Step 2. Let $r=3, y=\sqrt 2$, and $x=\sqrt {r^2-y^2}=\sqrt 7$. Step 3. Sides $x,y,r$ form a right triangle with angle $t$ facing $y$. Step 4. We have $cos(t)=\frac{x}{r}=\frac{\sqrt 7}{3}$ or $cos(sin^{-1}\frac{\sqrt 2}{3})=\frac{\sqrt 7}{3}$
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