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: 39


$tan~x = \frac{2}{\sqrt{5}+1}$

Work Step by Step

If $\pi \lt x \lt \frac{3\pi}{2}$, then the angle $x$ is in quadrant III. Then $2\pi \lt 2x \lt 3\pi$, so angle $2x$ is in quadrant I or quadrant II. Since $tan~2x$ is positive, then angle $2x$ is in quadrant I. $tan~2x = 2 = \frac{opp}{adj}$. If the adjacent side is 1 and the opposite side is 2, then the hypotenuse is $\sqrt{2^2+1^2} = \sqrt{5}$ We can find the value of $tan~x$: $tan~x = \frac{sin~2x}{1+cos~2x}$ $tan~x = \frac{\frac{2}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}$ $tan~x = \frac{\frac{2}{\sqrt{5}}}{\frac{\sqrt{5}+1}{\sqrt{5}}}$ $tan~x = \frac{2}{\sqrt{5}+1}$
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