Question
Prove that the function f given by f(x) = logsin is strictly increasing on $\Big(0,\ \frac{\pi}{2}\Big)$ and strictly decreasing on $\Big(\frac{\pi}{2},\ \pi\Big).$

Answer

Given: $\text{f}\text{(x)} = \log\sin \text{x} \ \Rightarrow\ \text{f}'\text{(x)}=\frac{1}{\sin \text{x}}\frac{\text{d}}{\text{dx}}\sin \text{x}$ $=\frac{1}{\sin \text{x}}\cos \text{x}=\cot \text{x}$
On the interva $\Big(0,\ \frac{\pi}{2}\Big)$ i.e., in first quadrant, f'(x) = cot x > 0
Therefore, f(x) is strictly increasing on $\Big(0,\ \frac{\pi}{2}\Big).$
On the interval $\Big(\frac{\pi}{2},\ \pi\Big)$ i.e., in second quadrant, f'(x) = cot x< 0
Therefore, f(x) is strictly decreasing on $\Big(\frac{\pi}{2},\ \pi\Big).$

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