Sum from $n=1$ to $N$ of $\sin(nx)/2^n$ The question states:

I'm not sure what to do here. I used the geometric sum to do $2^{-n}$, but I can't think of a way to sum $\sin(n\theta)$ to get the required RHS.

Further working 

Apologies for maybe being slow or missing something obvious here... I don't know how to get it into required form.

The solution below is just to give a basic proof without using Euler's identity. 
\begin{align*}
&\quad \sum_{n=1}^N \frac{\sin(n\theta)}{2^n} \cdot 2^N(5-4\cos\theta) \\
&=  \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right)(1+4-4\cos \theta) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right) + \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta \right) \\
&\quad -  \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta\cos\theta \right) \\
&=  \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right) +  \left( \sum_{n=1}^{N} 2^{n+1}\sin(N-n+1)\theta \right) \\
&\quad - \left( \sum_{n=1}^N 2^n [\sin(N-n+2)\theta + \sin(N-n)\theta] \right) \\
&= \left[ 2^{N-1}\sin\theta + \sum_{n=2}^{N-1}2^{n-1}\sin(N-n+1)\theta + \sin(N\theta) \right] \\
&\quad + \left[ 2^{N+1}\sin\theta + \sum_{n=2}^{N-1}2^{n+1}\sin(N-n+1)\theta + 2^2\sin(N\theta) \right] \\
&\quad - \left[ 2\sin(N+1)\theta + 2^2\sin(N\theta) + \sum_{n=2}^{N-1}2^{n+1}\sin(N-n+1)\theta + \sum_{n=2}^{N-1}2^{n-1}\sin(N-n+1)\theta + 2^{N-1}\sin\theta \right] \\
&= 2^{N+1}\sin\theta + \sin(N\theta) - 2\sin((N+1)\theta).
\end{align*}
But I do spend almost half an hour to write it, hah.
