Prove that the function $f$ given by $f(x) = \log(\sin x)$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

Given function is $f(x) = \log(\sin x)$.
First,we find the derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(\log(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Case $1$: In the interval $\left(0, \frac{\pi}{2}\right)$,the value of $\cot x$ is positive $(\cot x > 0)$.
Since $f'(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$,the function $f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.
Case $2$: In the interval $\left(\frac{\pi}{2}, \pi\right)$,the value of $\cot x$ is negative $(\cot x < 0)$.
Since $f'(x) < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$,the function $f(x)$ is strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$.