Question
Show that $\text{f}(\text{x})=\cos^2\text{x}$ is a decreasing function on $\Big(0,\frac{\pi}{2}\Big).$

Answer

We have,
$\text{f}(\text{x})=\cos^2\text{x}$
$\therefore\ \text{f}'(\text{x})=2\cos\text{x}(-\sin\text{x})$
$\Rightarrow\text{f}'(\text{x})=-2\sin\text{x}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-\sin2\text{x}$
Now,
$\text{x}\in\Big(0,\frac{\pi}{2}\Big)$
$\Rightarrow2\text{x}\in(0,\pi)$
$\Rightarrow\sin2\text{x}>0$ when $2\text{x}\in(0,\pi)$
$\Rightarrow-\sin2\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0$
So, f(x) is decreasing function on $\Big(0,\frac{\pi}{2}\Big).$

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