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

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

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:

It is given that $f(x)=\sin x$
$\begin{array}{l} \Rightarrow f(x)=\frac{d}{d x}(\sin x) \\ \Rightarrow f^{\prime}(x)=\cos x \end{array}$
Taking different region from 0 to $2 \pi$ we get
$\begin{array}{l} \text { Let }^{\mathrm{X} \in\left(0, \frac{\pi}{2}\right)} \\ {\Rightarrow \operatorname{Cos}(\mathrm{x})>0} \\ {\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0} \end{array}$
Therefore $f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$
$\begin{array}{l} \text { Let }^{x \in\left(\frac{\pi}{2}, \pi\right)} \\ \Rightarrow \operatorname{Cos}(x)<0 \\ \Rightarrow f^{\prime}(x)<0 \end{array}$
Therefore $f(x)$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$
Thus, from above condition we find that
$\Rightarrow \mathrm{f}(\mathrm{x})$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$
As a result, condition for $f(x)$ neither increasing nor decreasing in $(0, \pi)$