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: 33

Answer

See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression: $\sin \left( x+\frac{\pi }{2} \right)$ By using the identity of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta $, the above expression can be further simplified as: $\begin{align} & \sin \left( x+\frac{\pi }{2} \right)=\sin \,x\,\cos \,\frac{\pi }{2}+\cos \,x\,\sin \,\frac{\pi }{2} \\ & =\sin \,x\times 0+\cos \,x\times 1 \\ & =0+\cos \,x \\ & =\cos \,x \end{align}$ Thus, the left side of the expression is equal to the right side, which is $\sin \left( x+\frac{\pi }{2} \right)=\cos \,x$.
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