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

We have, $f(x) = \log (\sin x)$ …(i)

Differentiating (i) w.r.t. $x$, we get $f'(x) = \cfrac{1}{{\sin x}}(\cos x) = \cot x$

As $\cot x > 0$ for all $x \in \left( {0,\cfrac{\pi }{2}} \right)$ and $\cot x < 0$ for all $x \in \left( {\cfrac{\pi }{2},\pi } \right)$

Therefore, $f(x)$ is strictly increasing on $\left( {0,\cfrac{\pi }{2}} \right)$ and strictly decreasing on $\left( {\cfrac{\pi }{2},\pi } \right)$.