Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative $y'=1-cot^2\theta=2-csc^2\theta, 0\lt\theta\lt\pi$, we have $y''=2csc^2\theta\ cot\theta$
Step 2. Possible critical points are $y'=0, \theta=\frac{\pi}{4}, \frac{3\pi}{4}$. Determine the increasing $y'\gt0$ and decreasing $y'\lt0$ regions by signs of $y'$: $(0)..(-)..(\frac{\pi}{4})..(+)..(\frac{3\pi}{4})..(-)..(\pi)$; thus the function decreases on $(0,\frac{\pi}{4}), (\frac{\pi}{4}, \pi)$, increases on $(\frac{\pi}{4}, \frac{3\pi}{4})$. Local maximum at $\theta=\frac{3\pi}{4}$ and local minimum at $\theta=\frac{\pi}{4}$.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $\theta=\frac{\pi}{2}$: $(0)..(+)..(\frac{\pi}{2})..(-)..(\pi)$. Thus the function is concave up on $(0,\frac{\pi}{2})$ and concave dow on $(\frac{\pi}{2},\pi)$ with $\theta=\frac{\pi}{2}$ as the inflection point.
Step 4. Sketch the function based on the above results as shown.
