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: 48



Work Step by Step

$$\cos^2\frac{5\pi}{12}$$ *Recall the half-angle formula for cosine, which is $$\cos^2\theta=\frac{1+\cos2\theta}{2}$$ Thus, $$\cos^2\frac{5\pi}{12}=\frac{1+\cos\frac{5\pi}{6}}{2}$$ In the unit circle, angle $\frac{5\pi}{6}$ is the equivalent of angle $\frac{\pi}{6}$ but in the second quadrant, where cosine is negative. Therefore, $\cos\frac{5\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}$ So, return to the given function: $$\cos^2\frac{5\pi}{12}=\frac{1-\frac{\sqrt3}{2}}{2}$$ $$\cos^2\frac{5\pi}{12}=\frac{\frac{2-\sqrt3}{2}}{2}$$ $$\cos^2\frac{5\pi}{12}=\frac{2-\sqrt3}{4}$$
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