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: 41


$tan~\frac{x}{2} = 0.5$

Work Step by Step

If $0 \lt x \lt \frac{\pi}{2}$, then the angle $x$ is in quadrant I. If $sin~x = 0.8$, then $cos~x = \sqrt{1-sin^2~x} = 0.6$ We can find the value of $tan~\frac{x}{2}$: $tan~\frac{x}{2} = \frac{sin~x}{1+cos~x}$ $tan~\frac{x}{2} = \frac{0.8}{1+0.6}$ $tan~\frac{x}{2} = \frac{0.8}{1.6}$ $tan~\frac{x}{2} = 0.5$
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