Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 58

Answer

$$ \cos \frac{\pi}{8} =\sqrt{\frac{1}{2}+\frac{1}{2 \sqrt{2}}}$$

Work Step by Step

Since $$\cos2\theta =\frac{1}{2}\left(1+\cos 2\theta \right) $$ Then \begin{aligned} \cos ^{2} \frac{\pi / 4}{2} &=\frac{1+\cos \pi / 4}{2} \\ \cos ^{2} \frac{\pi}{8} &=\frac{1+\frac{1}{\sqrt{2}}}{2} \\ &=\frac{1}{2}+\frac{1}{2 \sqrt{2}} \end{aligned} Hence $$ \cos \frac{\pi}{8} =\sqrt{\frac{1}{2}+\frac{1}{2 \sqrt{2}}}$$
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