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 \((\frac{\pi}{2},\pi)\).