Show that the function given by $f\left( x \right) = \sin x$ is

(a) strictly increasing in $\left( {0,\cfrac{\pi }{2}} \right)$

(b) strictly decreasing in $\left( {\cfrac{\pi }{2},\;\pi } \right)$

(c) neither increasing nor decreasing in $\left( {0,\pi } \right)$

We have, $f(x) = \sin x$ …(i)
which is continuous and derivable on $R$

Differentiating (i) w.r.t. $x$, we get $f'(x) = \cos x$

(a) For all $x \in \left( {0,\cfrac{\pi }{2}} \right),\cos x > 0 \Rightarrow f'(x) > 0$

Therefore, $f(x) = \sin x$ is strictly increasing on $\left( {0,\cfrac{\pi }{2}} \right)$.

(b) For all $x \in \left( {\cfrac{\pi }{2},\pi } \right),\cos x < 0 \Rightarrow f\prime (x) < 0$
therefore $f(x) = \sin x$ is strictly decreasing on $\left( {\cfrac{\pi }{2},\pi } \right)$.

(c) From first two parts i.e. (a) and (b), we can conclude that $f\left( x \right) = \sin x$ is neither increasing nor decreasing on $(0,\;\pi )$ .