 # Show that f(x)=sin x is increasing `
Question:

Show that $f(x)=\sin x$ is increasing on $(0, \pi / 2)$ and decreasing on $(\pi / 2, \pi)$ and neither increasing nor decreasing in $(0, \pi)$.

Solution:

Given:- Function $f(x)=\sin x$

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 $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

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,

$f(x)=\sin x$

$\Rightarrow f(x)=\frac{d}{d x}(\sin x)$

$\Rightarrow f^{\prime}(x)=\cos x$

Taking different region from 0 to $2 \pi$

a) let $x \in\left(0, \frac{\pi}{2}\right)$

$\Rightarrow \cos (x)>0$

$\Rightarrow f^{\prime}(x)>0$

Thus $f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$

b) let $x \in\left(\frac{\pi}{2}, \pi\right)$

$\Rightarrow \cos (x)<0$

$\Rightarrow f^{\prime}(x)<0$

Thus $f(x)$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$

Therefore, from above condition we find that

$\Rightarrow f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$

Hence, condition for $f(x)$ neither increasing nor decreasing in $(0, \pi)$