Answer
The absolute maximum is $f(\frac{\pi}{2}) = 1$
The absolute minimum is $f(-\frac{\pi}{2}) = -1$
There is no local maximum/minimum.
Work Step by Step
$f(x) = sin~x$, $~~-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
The absolute maximum is $f(\frac{\pi}{2}) = 1$
The absolute minimum is $f(-\frac{\pi}{2}) = -1$