Answer
See below:
Work Step by Step
(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 $\cos x$.
