## Precalculus (6th Edition) Blitzer

(b) The relation between the two equations can be validated for equivalency. In order to verify the equations, use the trigonometric identity. $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ By using the above identity in the following equation, we get: $\sin \left( x-\frac{3\pi }{2} \right)=\sin x\cos \frac{3\pi }{2}-\cos x\sin \frac{3\pi }{2}$ Apply the values $\cos \frac{3\pi }{2}=0$ and $\sin \frac{3\pi }{2}=-1$ \begin{align} & \sin \left( x-\frac{3\pi }{2} \right)=\sin x\cos \frac{3\pi }{2}-\cos x\sin \frac{3\pi }{2} \\ & =\sin x.\left( 0 \right)-\cos x.\left( -1 \right) \\ & =0+\cos x \\ & =\cos x \end{align} Hence, the equation $\sin \left( x-\frac{3\pi }{2} \right)$ is equal to $\cos x$.