 Solution

Within the requested interval, the function is symmetric about $x=\pi /2$ because $\sin (\pi -x)=\sin x$. Its derivative $\cot x$ is positive before $\pi /2$, zero there, and negative afterwards. The negative second derivative shows that this is a strictly concave function. Consequently its unique maximum is $(\pi /2,0)$. It tends to $-\infty$ at both ends, giving vertical asymptotes $x=0$ and $x=\pi$.