Question

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

Answer 1

$$\begin{gathered} f\left(x\right)=\log \cos x \\ f\text{'}\left(x\right)=\frac{1}{\cos x}\left(-\sin x\right) \\ =-\tan x \\ \mathrm{Now}, \\ x\in \left(-\frac{\mathrm{\pi }}{2},0\right) \\ \Rightarrow \tan x<0 \\ \Rightarrow -\tan x>0 \\ \Rightarrow f\text{'}\left(x\right)>0 \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is strictly increasing on}\left(-\frac{\mathrm{\pi }}{2},0\right). \\ \mathrm{Now}, \\ x\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \tan x>0 \\ \Rightarrow -\tan x<0 \\ \Rightarrow f\text{'}(x)<0 \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is}\mathrm{strictly}\text{decreasing on}\left(0,\frac{\mathrm{\pi }}{2}\right)\text{.} \end{gathered}$$