Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 29

Answer

$-\frac{\sqrt {14}}{2}$.

Work Step by Step

1. Let $sin^{-1}(-\frac{\sqrt 2}{3})=t$, we have $sin(t)=-\frac{\sqrt 2}{3}$ and $t$ in quadrant IV. 2. Let $r=3, y=-\sqrt 2$, we have $x=\sqrt {3^2-(-\sqrt 2)^2}=\sqrt 7$. 3. Thus $cot(sin^{-1}(-\frac{\sqrt 2}{3}))=cot(t)=\frac{x}{y}=-\frac{\sqrt {14}}{2}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.