Trig Stuff 2

Note that $\displaystyle S = \sum_{j=-5}^5 \left(\sum_{i=-5}^{j-1} \sin \frac {i \pi}{11} \sin \frac {j \pi}{11} \right) = \sum_{\substack {i,j=-5 \\ i \ne j}}^n \sin \frac {i \pi}{11} \sin \frac {j \pi}{11}$ and that:

\begin{aligned} \left(\sum_{i=-5}^5 \sin \frac {i \pi}{11} \right)^2 & = \sum_{i=-5}^5 \sin^2 \frac {i \pi}{11} - 2 \sum_{\substack {i,j=-5 \\ i \ne j}}^n \sin \frac {i \pi}{11} \sin \frac {j \pi}{11} \\ \implies S & = \frac 12 \left(\left({\color{#3D99F6}\sum_{i=-5}^5 \sin \frac {i \pi}{11}} \right)^2 - \sum_{i=-5}^5 \sin^2 \frac {i \pi}{11} \right) & \small \color{#3D99F6} \text{Since }\sin (-x) = - \sin x \text{ and } \sin 0 = 0 \\ & = \frac 12 \left({\color{#3D99F6}0} - 2 \sum_{i=1}^5 \sin^2 \frac {i \pi}{11} \right) \\ & = - \sum_{i=1}^5 \frac 12 \left(1-\cos \frac {2i \pi}{11} \right) \\ & = - \frac 12 \left(\sum_{i=1}^5 1 - {\color{#3D99F6} \sum_{i=1}^5 \cos \frac {2i \pi}{11}} \right) & \small \color{#3D99F6} \text{Note that } \sum_{k=1}^n \cos \frac {2k \pi}{2n+1} = - \frac 12 \\ & = - \frac 12 \left(5- \left({\color{#3D99F6} -\frac 12}\right) \right) \\ & = - \frac {11}4 \end{aligned}

$\implies a+b = 11+4 = \boxed{15}$

Hint: Vieta's formulas.