Show that the function given by $f(x)=\sin x$ is  (a) strictly increasing in $\left(0,\frac{\pi }{2}\right)$  (b) strictly decreasing in $\left(\frac{\pi }{2},\pi \right)$ (c) neither increasing nor decreasing in $(0,\pi )$.

The given function is  $f(x)=\sin x.$
 $\therefore f^{\prime }(x)=\cos x$
(a) Since for each $x\in \left(0,\frac{\pi }{2},\right),\cos x>0\Rightarrow f^{\prime }(x)>0.$
Hence, $f$ is strictly increasing in $\left(0,\frac{\pi }{2}\right)$.
(b) Since for each $x\in \left(\frac{\pi }{2},\pi \right),\cos x<0\Rightarrow f^{\prime }(x)<0$.
Hence, $f$ is strictly decreasing in $\left(\frac{\pi }{2},\pi \right)$.
(c) From the results obtained in (a) and (b), it is clear that $f$ is neither increasing nor
decreasing in $(0,\pi )$.