Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 249: 57


$tan~\theta~sin~2\theta = 2-2~cos^2~\theta$

Work Step by Step

$tan~\theta~sin~2\theta = \frac{sin~\theta}{cos~\theta}~(2~sin~\theta~cos~\theta)$ $tan~\theta~sin~2\theta = 2~sin^2~\theta$ $tan~\theta~sin~2\theta = 2~(1-cos^2~\theta)$ $tan~\theta~sin~2\theta = 2-2~cos^2~\theta$
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