Sum of geometric series where $x=\sin^{-1}{\frac{7}{8}}$

Question

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, $\sin{2x}$?

Answer 1

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 $$ c=2\cos x=2\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{4}<1 $$

Answer 2

The series seems to be

$$S=\sin(x)\sum_{n=0}^{\infty}(2\cos(x))^n$$

Since $x=\arcsin(7/8)$, $2\cos(x)=\sqrt{15}/4<1$. So the series converges and

$$S=(7/8)\frac1{1-\sqrt{15}/4}$$