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: 49



Work Step by Step

$$\sin^2\frac{\pi}{12}$$ *Recall the half-angle formula for sine, which is $$\sin^2\theta=\frac{1-\cos2\theta}{2}$$ Thus, $$\sin^2\frac{\pi}{12}=\frac{1-\cos\frac{\pi}{6}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{1-\frac{\sqrt3}{2}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{\frac{2-\sqrt3}{2}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{2-\sqrt3}{4}$$
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