Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 52


See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression, $\cos \,2\alpha $ By using the identity of trigonometry, $\cos \,\left( \alpha +\beta \right)=\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta $ Now, the above expression can be further simplified as, $\begin{align} & \cos \,2\alpha =\cos \,\left( \alpha +\alpha \right) \\ & =\cos \,\alpha \,\cos \,\alpha -\sin \,\alpha \,\sin \,\alpha \\ & ={{\cos }^{2}}\,\alpha -{{\sin }^{2}}\,\alpha \end{align}$ Thus, the left side of the given expression is equal to the right side, which is $\cos \,2\alpha ={{\cos }^{2}}\,\alpha -{{\sin }^{2}}\,\alpha $.
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