Answer
The absolute maximum $f(\pi/2)=1$ and the absolute minimum $f(-\pi/2)=-1$.
See graph.
Work Step by Step
Step 1. Given the function $f(\theta)=sin\theta, -\pi/2\leq \theta\leq 5\pi/6$, we have $f(-\pi/2)=-1, f(5\pi/6)=1/2$,
Step 2. $f'(\theta)=cos\theta$, let $f'(\theta)=0$, we get critical points at $\theta=-\pi/2, \pi/2$ and $f(\pi/2)=1$
Step 3. We can identify the absolute maximum as $f(\pi/2)=1$ and the absolute minimum as $f(-\pi/2)=-1$ on the given interval.
Step 4. See graph.