Precalculus (6th Edition) Blitzer

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

Chapter 5 - Mid-Chapter Check Point - Page 684: 24


The required value is $\frac{\sqrt{3}}{2}$

Work Step by Step

We have the double angle formulas, ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $. Therefore, applying the formula in the given expression where $\theta ={{15}^{\circ }}$, we get: $\begin{align} & {{\cos }^{2}}15-{{\sin }^{2}}15=\cos \left( {{2.15}^{\circ }} \right) \\ & =\cos \left( {{30}^{\circ }} \right) \\ & =\frac{\sqrt{3}}{2} \end{align}$ Hence, the required value is $\frac{\sqrt{3}}{2}$
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