Sum of geometric series where $x=\sin^{-1}{\frac{7}{8}}$ This problem is from a local contest. 
The series $\sin{x}$, $\sin{2x}$, $4\sin{x}-4\sin^3{x}$, $...$ is a geometric series if $x=\sin^{-1}{\frac{7}{8}}$. Compute the sum of the geometric series.
I did not compute the solution in time but this was my reasoning. 
For the first term, $\sin{\sin^{-1}{\frac{7}{8}}}=\frac{7}{8}$
I was not able to find $\sin{2x}$. I did not attempt the double angle identity since that would have introduced cosine.
For the third term, I found $4\sin{\sin^{-1}{\frac{7}{8}}}=\frac{28}{8}$ and $4\sin^3{\sin^{-1}{\frac{7}{8}}}=4(\frac{7}{8})^3=\frac{1372}{512}$. The third term is the difference of $\frac{28}{8}$ and $\frac{1372}{512}$. 
My question is, how can I compute the second term, $\sin{2x}$?

A sequence $(a_n)$ is a geometric series if $a_{n+1}/a_n=c$, for every $n$. Thus the first two terms suffice to determine it:
$$
a_0=\sin x,\qquad a_1=\sin2x=2\sin x\cos x
$$
Then
$$
c=\frac{a_1}{a_0}=2\cos x
$$
and indeed
$$
a_2=4\sin x-4\sin^3x=4\sin x\cos^2x=a_1\cdot 2\cos x
$$
Thus you have
$$
a_n=2^n\cos^nx\sin x
$$
and the sum of a geometric series converges if and only if $|c|<1$.
In your case
$$
c=2\cos x=2\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{4}<1
$$
