Question

Show that $f\left(x\right)=\sin x$ is increasing on $(0,\frac{\pi }{2})$ and decreasing on $(\frac{\pi }{2},\pi )$ and neither inreasing nor decreasing in $(0,\pi )$.

Answer 1

$f\left(x\right)=\sin x$

$f^{\prime }\left(x\right)=\cos x$

(i) for increasing function

$f^{\prime }\left(x\right)>0$ at $(0,\frac{\pi }{2})$

$\cos x>0$ at $(0,\frac{\pi }{2})$

$\therefore$ The function $f\left(x\right)$ is increasing

(ii) for decreasing function

$f^{\prime }\left(x\right)<0$ at $(\frac{\pi }{2},\pi )$

$\therefore$ The function $f\left(x\right)$ is decreasing

(iii) at $(0,\pi )$

$f^{\prime }\left(x\right)=\cos x$

$f^{\prime }\left(0\right)=\cos \left(0\right)=1$

$f^{\prime }\left(\pi \right)=\cos \pi =-1$

Hence the function $f\left(x\right)$ neither decrease nor increases.