## Precalculus (6th Edition) Blitzer

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 $\sin \left( x+\frac{\pi }{2} \right)=\cos \,x$.