Question
Show that $\text{f}(\text{x})=\log\sin\text{x}$ is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and decreasing on $\Big(\frac{\pi}{2},\pi\Big).$

Answer

$\text{f}(\text{x})=\log\sin\text{x}$
$\text{f}'(\text{x})=\frac{1}{\sin\text{x}}\cos\text{x}=\cot\text{x}$
In interval $\Big(0,\frac{\pi}{2}\Big),\text{f}'(\text{x})=\cot\text{x}>0.$
$\therefore$ f is strictly increasing in $\Big(0,\frac{\pi}{2}\Big).$
In interval $\Big(\frac{\pi}{2},\pi\Big),\text{f}'(\text{x})=\cot\text{x}<0.$
$\therefore$ f is strictly decreasing in $\Big(\frac{\pi}{2},\pi\Big).$

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