Question

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

Answer 1

We have,

f(x)=log cos x

\(fx=\frac {1}{cosx} (-sinx)=-tan x\)

In interval \((0,\frac{\pi}{2})\), tan x>0=- tanx<0

f'(x)<0 on \((0,\frac{\pi}{2})\)

∴ f is strictly decreasing in \((0,\frac{\pi}{2})\)

In interval \((\frac{\pi}{2},\pi)\),tan x<0=- tanx>0

f'(x)>0 on \((\frac{\pi}{2},\pi)\)

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