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

Question

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

✓

Answer

Given : $f(x)=\cos ^2 x$
Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a,b)$.
i. If $f^{\prime}(x);0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
ii. If $f^{\prime}(x)<0$ for all,$x \in(a, b)$ then $f(x)$ is decreasing on $(a, b)$
For the value of $x$ obtained in $(ii)\ f(x)$ is increasing and for remaining points in its domain it is decreasing.
Here we have,
$ f ( x )=\cos ^2 x$
$\Rightarrow f(x)=\frac{d}{d x}\left(\cos ^2 x\right)$
$= f ^{\prime}( x )=3 \cos x (-\sin x )$
$= f ^{\prime}( x )=-2 \sin ( x ) \cos ( x )$
$= f ^{\prime}( x )=-\sin 2 x $ ; as $ \sin 2 A=2 \sin A \cos A $
Now, as given
$x \in\left(0, \frac{\pi}{2}\right)$
$=2 x \in(0, \pi)$
$=\operatorname{Sin}(2 x);0$
$=-\operatorname{Sin}(2 x)<0$
$\Rightarrow f^{\prime}(x)<0$
hence, it is the condition for $f(x)$ to be decreasing
Thus $,f(x)$ is decreasing on interval $\left(0, \frac{\pi}{2}\right)$