Question:
Prove that the function $f$ given by $f(x)=\log \sin x$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ and strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$.
Solution:
We have,
$f(x)=\log \sin x$
$\therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=\cot x>0$.
$\therefore f$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.
In interval $\left(\frac{\pi}{2}, \pi\right), f^{\prime}(x)=\cot x<0$
$\therefore f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.
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