Explanation
We are given the function:
[tex]f(x)=\sin x[/tex]
We are required to determine the equivalent trigonometric function as that given above.
We know that the following trigonometric equivalence exists:
[tex]\begin{gathered} \cos(90-x)=\sin x \\ \cos(90+x)=-\sin x \\ \cos(180-x)=-\cos x \\ \cos(180+x)=-\cos x \\ \cos(270-x)=-\sin x \\ \cos(270+x)=\sin x \\ \cos(360-x)=\cos x \\ \cos(360+x)=\cos x \end{gathered}[/tex]
Next, we determine the value of each option as follows:
[tex]\begin{gathered} Option\text{ }A:f(x)=\cos(x-\frac{3\pi}{2})=\cos\lbrace-(\frac{3\pi}{2}-x)\rbrace \\ =\cos(\frac{3\pi}{2}-x)=\cos(270-x)=-\sin x \\ \\ Option\text{ }B:f(x)=\cos(x-\frac{\pi}{2})=\cos\lbrace-(\frac{\pi}{2}-x)\rbrace \\ =\cos(\frac{\pi}{2}-x)=\cos(90-x)=\sin x \\ \\ Option\text{ }C:f(x)=\cos(-x-\frac{\pi}{2})=\cos\lbrace-(\frac{\pi}{2}+x)\rbrace \\ =\cos(\frac{\pi}{2}+x)=\cos(90+x)=-\sin x \\ \\ Option\text{ }D:f(x)=\cos(x+\pi)=\cos(180+x)=-\cos x \end{gathered}[/tex]
Hence, the answer is option B.
[tex]\begin{equation*} f(x)=\cos(x-\frac{\pi}{2}) \end{equation*}[/tex]
