Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 4

Answer

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Work Step by Step

$cos~\frac{5\pi}{12}$ $=cos~(\frac{3\pi}{12}+\frac{2\pi}{12})$ $=cos~(\frac{\pi}{4}+\frac{\pi}{6})$ $=cos~\frac{\pi}{4}~cos~\frac{\pi}{6} - sin~\frac{\pi}{4}~sin~\frac{\pi}{6}$ $=(\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$ $=\frac{\sqrt{6}-\sqrt{2}}{4}$
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