Answer
$\frac{1}{\sqrt {u^2+1}}$
Work Step by Step
1. Let $cot^{-1}u=t, 0\lt t\lt\pi$, we have $cot(t)=u$.
2. Let $x=u,y=1$, we have $r=\sqrt {u^2+1}$
3. Thus $sin(cot^{-1}u)=sin(t)=\frac{y}{r}=\frac{1}{\sqrt {u^2+1}}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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 481: 62
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