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.
 A: 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.
A: HINT:
Euler's identity says: $$e^{ix}=\cos x+i\sin x$$
$\implies\dfrac{\sin n\theta}{2^n}=$ imaginary part of $\left(\dfrac{e^{i\theta}}2\right)^n$
A: $$e^{inx}=\cos(nx)+i\sin(nx),$$
yet $$e^{-inx}=\cos(nx)-i\sin(nx)$$
so we can deduce that $$\sin(x)=\frac{e^{inx}-e^{-inx}}{2i}.$$
This is annoying, but will get the job done. See Dirichlet Kernel, which uses a similar method, and is somewhat related (at least you know this is not a meaningless computation.)
The hint was intended to suggest that you wanted something of the form $a\cdot r^n,$ and in this case, you are summing over the sum of two power series:
$$\sum_{n=1}^{N}\frac{e^{inx}-e^{-inx}}{2^{n+1}i}.$$

Edit:
\begin{align*}\sum_{n=1}^{N}\frac{e^{inx}-e^{-inx}}{2^{n+1}i}&=\sum_{n=1}^{N}\frac{e^{inx}}{2^{n+1}i}-\sum_{n=1}^{N}\frac{e^{-inx}}{2^{n+1}i}\\
&=\frac{1}{2i}\left(  \sum_{n=1}^{N}\frac{e^{inx}}{2^n}-\sum_{n=1}^{N}\frac{e^{-inx}}{2^{n}}\right)\\
&=\frac{1}{2i}\left(  \sum_{n=1}^{N}\left(\frac{e^{ix}}{2}\right)^n-\sum_{n=1}^{N}\left(\frac{e^{-ix}}{2}\right)^n\right)
\end{align*}
'
Use the hint, along with my comment. Notice what "$r$" is, and see if you can finish the problem.
