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 - Cumulative Review - Page 529: 9

Answer

$\frac{\sqrt 5}{5}$

Work Step by Step

Let $tan^{-1}2=t$, we have $tan(t)=2$. Set $x=1, y=2$, then $r=\sqrt {1^2+2^2}=\sqrt 5$ and $cos(t)=\frac{x}{r}=\frac{\sqrt 5}{5}$
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