Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 20


The exact value of $1-2{{\sin }^{2}}\frac{\pi }{12}$ is $\frac{\sqrt{3}}{2}$.

Work Step by Step

Recall the given expression. $\cos 2\theta =1-2{{\sin }^{2}}\theta $ Apply the given expression. $\begin{align} & 1-2{{\sin }^{2}}\frac{\pi }{12}=\cos 2\left( \frac{\pi }{12} \right) \\ & =\cos \left( \frac{\pi }{6} \right) \\ & =\frac{\sqrt{3}}{2} \end{align}$ Therefore, the exact value of $1-2{{\sin }^{2}}\frac{\pi }{12}$ is $\frac{\sqrt{3}}{2}$.
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