Question

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

Answer 1

We have,

f(x)=log sin x

f'(x)=$\frac{1}{sinx}$ cosx=cot x

In interval $(0,\frac{\pi}{2})$ f'(x)=cot x>0.

∴ f is strictly increasing in $(0,\frac{\pi}{2})$.

In interval $(\frac{\pi}{2},\pi)$, f'(x)=cot x<0.

∴f is strictly decreasing in $(\frac{\pi}{2},\pi)$.