How can I prove that $\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$? $$\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$$
Have no clue how to prove it.
Possibly rewrite $\sin(7n)$ as $\frac{1}{2\sin(7)}\left(\cos(7n-7)-\cos(7n+7)\right)$.
But what next?
 A: Here is a method which does not use complex numbers. Let $x=\sum_{n=0}^\infty\frac{\sin(7n)}{13^n}$ and $y=\sum_{n=0}^\infty\frac{\cos(7n)}{13^n}$.
Then, we have
\begin{align}
x&=\sum_{n=1}^\infty\frac{\sin(7n)}{13^n}=\sum_{n=0}^\infty\frac{\sin(7n+7)}{13^{n+1}}\\
&=\frac1{13}\sum_{n=0}^\infty\frac{\sin(7)\cos(7n)+\cos(7)\sin(7n)}{13^n}=\frac{\sin(7)}{13}y+\frac{\cos(7)}{13}x.
\end{align}
Similarly,
\begin{align}
y&=1+\sum_{n=1}^\infty\frac{\cos(7n)}{13^n}=1+\sum_{n=0}^\infty\frac{\cos(7n+7)}{13^{n+1}}\\
&=1+\frac1{13}\sum_{n=0}^\infty\frac{\cos(7n)\cos(7)-\sin(7n)\sin(7)}{13^n}=1+\frac{\cos(7)}{13}y-\frac{\sin(7)}{13}x.
\end{align}
Now, all that is left is to solve for $x$.
A: To follow up on my comment: using the geometric series we have
$$
\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n} = \Im\left(\sum_{n=0}^{\infty}\left(\frac{e^{7i}}{13}\right)^n\right)
$$
$$
=\Im\left(\frac{13}{13-e^{7i}}\right)
$$Multiply by the conjugate and take the imaginary part.
$$
=\Im\left(\frac{13(13-e^{-7i})}{13^2+1-26\cos(7)}\right)
$$
$$
=\frac{13\sin(7)}{170-26\cos(7)}
$$
