Question

Show that f(x) = cos² x is a decreasing function on (0, π/2).

Answer 1

$$\begin{gathered} f\left(x\right)={\cos }^{2}x \\ f\text{'}\left(x\right)=2\cos x\left(-\sin x\right) \\ \Rightarrow f\text{'}\left(x\right)=-\sin \left(2x\right)...\left(1\right) \\ \text{Now,} \\ 0<x<\frac{\mathrm{\pi }}{2} \\ \Rightarrow 0<2x<\mathrm{\pi } \\ \Rightarrow \sin 2\mathrm{x}>0\left[\because \mathrm{Sine}\mathrm{fuction}\mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{quadrant}\right] \\ \Rightarrow -\sin 2\mathrm{x}<0 \\ \Rightarrow f\text{'}\left(\mathrm{x}\right)<0\left[\text{From eq. (1)}\right] \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(0,\frac{\mathrm{\pi }}{2}\right)\text{.} \end{gathered}$$