Finding values that make the series converge For which values of $\theta \in [0,2\pi)$ does the sum converge? And then for these values of $\theta $, find the sum of the series.
The given series for this question is $\sum_{n=0}^{\infty} (sin\theta)^n$  
So this particular series is a geometric series, and geometric series converge when $r<1$ and converge to $\frac{a}{1-r}$ and diverge when $r>=1$. 
So referring to the unit circle wouldn't all possible values of $\theta$ include $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3},\frac{2\pi}{3},\frac{3\pi}{4}, \frac{5\pi}{6}, \pi,  \frac{7\pi}{6}, \frac{5\pi}{4}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{4},\frac{11\pi}{6}  $ is this logic right? 
so then a would be 1 for this equation and r would be all of these values listed above?

The geometric series $\sum_{n=0}^{\infty} x^n$ converges if and only if $|x|<1$. You have noted that for the most part.
When is $|\sin (\theta)|<1$? Almost always except when $|\sin (\theta)|=1$. So it diverges only when $\sin (\theta)=1$ or $\sin (\theta)=-1$. There are a whole infinite amount of $\theta \in [0,2\pi]$ not just the ones you've listed.
