Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 528: 47

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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