Answer
$f(\theta)$ has the absolute maximum value of $1$ at $\theta=\pi/2$ and the absolute minimum value of $-1$ at $\theta=-\pi/2$ in the defined domain.
Work Step by Step
$$f(\theta)=\sin\theta\hspace{1cm}-\frac{\pi}{2}\le\theta\le\frac{5\pi}{6}$$
1) Find the derivative $f'(\theta)$: $$f'(\theta)=\cos\theta$$
We would have $f'(\theta)=0$ when $\theta=\pi/2+k\pi$ for $k\in Z$.
However, in the defined domain $[-\pi/2,5\pi/6]$, we accept only 2 values of $\theta$, which are $\theta=\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$. These are the critical points of $f(\theta)$.
2) Evaluate $f'(\theta)$ at the critical points and endpoints:
- For $\theta=-\pi/2$: $$f\Big(-\frac{\pi}{2}\Big)=\sin\Big(-\frac{\pi}{2}\Big)=-1$$
- For $\theta=\pi/2$: $$f\Big(\frac{\pi}{2}\Big)=\sin\Big(\frac{\pi}{2}\Big)=1$$
- For $\theta=5\pi/6$: $$f\Big(\frac{5\pi}{6}\Big)=\sin\Big(\frac{5\pi}{6}\Big)=\frac{1}{2}$$
This means $f(\theta)$ has the absolute maximum value of $1$ at $\theta=\pi/2$ and the absolute minimum value of $-1$ at $\theta=-\pi/2$ in the defined domain.