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 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 33



Work Step by Step

We have: $\sin \theta=\dfrac{2}{\sqrt 5}$ and $\cos \theta=-\dfrac{1}{\sqrt 5}$ Since, $\tan (\theta)=\dfrac{\sin \theta}{\cos \theta} \implies \tan \theta=\dfrac{2/\sqrt 5}{-1/\sqrt 5}=-2$ We plug these values into the following identity: $\tan (2 \ \theta)=\dfrac{2 \tan \theta}{ 1-\tan^2 \theta} = \dfrac{(2)(-2) }{ 1- (-2)^2}=\dfrac{4}{3}$
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