University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 47

Answer

$$\cos^2\frac{\pi}{8}=\frac{2+\sqrt2}{4}$$

Work Step by Step

$$\cos^2\frac{\pi}{8}$$ *Recall the half-angle formula for cosine, which is $$\cos^2\theta=\frac{1+\cos2\theta}{2}$$ Thus, $$\cos^2\frac{\pi}{8}=\frac{1+\cos\frac{\pi}{4}}{2}$$ $$\cos^2\frac{\pi}{8}=\frac{1+\frac{\sqrt2}{2}}{2}$$ $$\cos^2\frac{\pi}{8}=\frac{\frac{2+\sqrt2}{2}}{2}$$ $$\cos^2\frac{\pi}{8}=\frac{2+\sqrt2}{4}$$
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