Answer
$$\sin\Big(x-\frac{\pi}{2}\Big)=\cos x$$
The statement is false.
Work Step by Step
$$\sin\Big(x-\frac{\pi}{2}\Big)=\cos x$$
Now what we've already known from the cofunction identities is
$$\sin\Big(\frac{\pi}{2}-x\Big)=\cos x$$
Yet here, the statement involves $\sin\Big(x-\frac{\pi}{2}\Big)$, not $\sin\Big(\frac{\pi}{2}-x\Big)$. Unfortunately, 2 expressions are not the same, so we need to find a way to transform $\sin\Big(x-\frac{\pi}{2}\Big)$ to the already known.
And we can do like this:
$$\sin\Big(x-\frac{\pi}{2}\Big)=\sin\Big[-\Big(\frac{\pi}{2}-x\Big)\Big]$$
And we already know from the negative-angle identities that
$$\sin(-\theta)=-\sin\theta$$
Therefore,
$$\sin\Big(x-\frac{\pi}{2}\Big)=-\sin\Big(\frac{\pi}{2}-x\Big)$$
$$\sin\Big(x-\frac{\pi}{2}\Big)=-\cos x$$
Since $\cos x\ne-\cos x$, $$\sin\Big(x-\frac{\pi}{2}\Big)\ne\cos x$$
The statement thus is false.
