Answer
$\sin(\frac{\pi}{2}-x)=\sin(\frac{\pi}{2}+x)$
Work Step by Step
First, let's simplify the left side:
$\sin(\frac{\pi}{2}-x)$
Expand using the subtraction formula for sine:
$=\sin\frac{\pi}{2}\cos x-\cos\frac{\pi}{2}\sin x$
Evaluate $\sin\frac{\pi}{2}$ and $\cos\frac{\pi}{2}$:
$=1*\cos x-0*\sin x$
$=\cos x$
Now let's simplify the right side:
$\sin(\frac{\pi}{2}+x)$
Expand using the addition formula for sine:
$=\sin\frac{\pi}{2}\cos x+\cos\frac{\pi}{2}\sin x$
Evaluate $\sin\frac{\pi}{2}$ and $\cos\frac{\pi}{2}$:
$=1*\cos x+0*\sin x$
Simplify:
$=\cos x$
Since both sides are equal to $\cos x$, the identity is proven.
