Show that the function given by $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.

(N/A) The given function is $f(x) = \sin x$.
We find the derivative: $f'(x) = \cos x$.
$(A)$ For $x \in (0, \frac{\pi}{2})$,$\cos x > 0$,which implies $f'(x) > 0$.
Therefore,$f(x)$ is strictly increasing in $(0, \frac{\pi}{2})$.
$(B)$ For $x \in (\frac{\pi}{2}, \pi)$,$\cos x < 0$,which implies $f'(x) < 0$.
Therefore,$f(x)$ is strictly decreasing in $(\frac{\pi}{2}, \pi)$.
$(C)$ Since the function is strictly increasing in one part of the interval $(0, \pi)$ and strictly decreasing in another part,it is neither increasing nor decreasing on the whole interval $(0, \pi)$.