How can I prove that $\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$?

Question

$$\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$$

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?

Answer 1

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. $$ =\Im\left(\frac{13(13-e^{-7i})}{13^2+1-26\cos(7)}\right) $$ $$ =\frac{13\sin(7)}{170-26\cos(7)} $$

Answer 2

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 $x$.