Question

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

Answer 1

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