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: 60


$2~cos^2~\theta-1 = \frac{1-tan^2~\theta}{1+tan^2~\theta}$

Work Step by Step

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