Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative $y'=csc^2\frac{\theta}{2}, 0\lt\theta\lt2\pi, 0\lt\frac{\theta}{2}\lt\pi$, we have $y''=2csc\frac{\theta}{2}(-csc\frac{\theta}{2}\ cot\frac{\theta}{2})/2=-csc^2\frac{\theta}{2}cot\frac{\theta}{2}$
Step 2. Determine the increasing $y'\gt0$ and decreasing $y'\lt0$ regions by signs of $y'$: $(0)..(+)..(2\pi)$; thus the function increases in $(0,2\pi)$. There are no critical points and no local extrema.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=\pi$: $(0)..(-)..(\pi)..(+)..(2\pi)$. Thus the function is concave down on $(0,\pi)$ and concave up on $(\pi,2\pi)$ with $x=\pi$ as the inflection point.
Step 4. Sketch the function based on the above results as shown.
