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: 65


$tan~4\theta = \frac{2~tan~2\theta}{2-sec^2~2\theta}$

Work Step by Step

We can verify that the equation is an identity: $tan~4\theta$ $= \frac{sin~4\theta}{cos~4\theta}$ $= \frac{2~sin~2\theta~cos~2\theta}{2~cos^2~2\theta-1}$ $= \frac{2~sin~2\theta~cos~2\theta}{2~cos^2~2\theta-1}\cdot \frac{\frac{1}{cos^2~2\theta}}{\frac{1}{cos^2~2\theta}}$ $= \frac{\frac{2~sin~2\theta~cos~2\theta}{cos^2~2\theta}}{\frac{2~cos^2~2\theta-1}{cos^2~2\theta}}$ $= \frac{\frac{2~sin~2\theta}{cos~2\theta}}{2-\frac{1}{cos^2~2\theta}}$ $= \frac{2~tan~2\theta}{2-sec^2~2\theta}$
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