## Precalculus (6th Edition) Blitzer

(b) In order to verify the equations, use the trigonometric identity. $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$ Now, apply the above identity in the following equation: $\cos \left( x+\frac{\pi }{2} \right)=\cos x\cos \frac{\pi }{2}-\sin x\sin \frac{\pi }{2}$ By using the values $\cos \frac{\pi }{2}=0$ and $\sin \frac{\pi }{2}=1$, we get: \begin{align} & \cos \left( x+\frac{\pi }{2} \right)=\cos x\cos \frac{\pi }{2}-\sin x\sin \frac{\pi }{2} \\ & =\cos x.\left( 0 \right)-\sin x.\left( 1 \right) \\ & =0-\sin x \\ & =-\sin x \end{align} Hence, the equation $\cos \left( x+\frac{\pi }{2} \right)$ is equal to $-\sin x$.