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: 52

Answer

$ \frac{\sqrt {2+\sqrt 2}}{2}$

Work Step by Step

$sin\frac{5\pi}{8}=sin\frac{5\pi/4}{2}=\sqrt {\frac{1-cos5\pi/4}{2}} ==\sqrt {\frac{1+\sqrt 2/2}{2}}=\frac{\sqrt {2+\sqrt 2}}{2}$
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