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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?

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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.

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    okay so then all the values I have listed should be correct then? I guess with your edit I will just use those values cause it is impossible to list them all – 2017-01-17
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    It is correct but you forgot an infinite amount that work @bjp409 and some that do not work. – 2017-01-17
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    it would be impossible to include all of the possible values and to find their sums – 2017-01-17
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    As you noted $a=1$ and $r=1$ so the sum if converges is $\frac{1}{1-\sin (\theta)}$. To refer to the set of all $\theta$ that work just use set builder notation $S=\{\theta \in \mathbb{R} | \theta \neq \frac{\pi}{2}+\pi n, n \in \mathbb{Z} \}$. – 2017-01-17
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    @bjp409 not at all! Just list those that do *not* work, and say all the others do! – 2017-01-17