How would one evaluate $\sin(72\pi/11)$? How would one evaluate $\sin(\frac {72\pi} {11})$?. The prime number in the bottom is getting me stuck. I couldn't see how to use it using the sum of two angles trig identity. 

First of all, 
$$\sin{\left ( \frac{72 \pi}{11} \right )} = \sin{\left ( \frac{6 \pi}{11} \right )} = \cos{\left ( \frac{\pi}{22} \right )}$$
Now, let $y=\pi/11$.  Then
$$\sin{6 y} = \sin{5 y}$$
Expanding, we find that $\cos{y}$ is the real root of the following equation closest to $1$:
$$32 \cos^5{y} - 16 \cos^4{y} - 32 \cos^3{y} + 12 \cos^2{y}+6 \cos{y} -1=0$$
The quantity desired is then
$$\sin{\left ( \frac{72 \pi}{11} \right )} = \sqrt{\frac{1+\cos{y}}{2}} $$
Of course, this evaluation depends on obtaining a closed form for the root of a fifth-degree equation, but that's about as much as you can ask for in this case.
