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: 50


$2cos^3~x-cos~x = \frac{cos^2~x-sin^2~x}{sec~x}$

Work Step by Step

$2cos^3~x-cos~x = (2cos^2~x-1)(cos~x)$ $2cos^3~x-cos~x = \frac{2cos^2~x-(sin^2~x+cos^2~x)}{\frac{1}{cos~x}}$ $2cos^3~x-cos~x = \frac{cos^2~x-sin^2~x}{sec~x}$
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