Question

Prove that the function f given by f(x) = log cos x is strictly decreasing on[0,π/2] and strictly increasing on[π/2,π]

Answer 1

given, f(x) = logcosx
differentiate f(x) with respect to x,
f'(x) = d{logcosx}/dx = 1/cosx × d(cosx)/dx
= 1/cosx × (-sinx)
= -sinx/cosx = -tanx
hence, f'(x) = -tanx

in interval [0, π/2] , tanx > 0
so, -tanx = f'(x) < 0
therefore, f is strictly decreasing on [0, π/2]

in interval [π/2 , π] , tanx < 0
so, -tanx = f'(x) > 0
therefore, f is strictly increasing on [π/2, π]

$\textbf{\underline{hence proved}}$