Prove This using De Moivre Theorem
$$\sum_{n=1}^{N}\frac{\sin n\theta}{2^n}=\frac{2^{N+1}\sin\theta+\sin N\theta-2\sin(N+1)\theta}{2^N(5-4\cos\theta)}$$
Please help me find my mistake, because I am not getting this result.
What I did:
$$\frac{sin n\theta} {2^N} = \Im \frac{e^{n\theta i}}{2^N}$$
Applied the G.P sum formula :
$$\frac{\frac{e^{\theta i}}{2} (1-\frac{e^{N\theta i}}{2^N})}{1-\frac{e^{\theta i}}{2}}$$
$$\Im\frac{e^{\theta i} (\frac{2^N-e^{N\theta i}}{2^N})}{2-{e^{\theta i}}}$$
$$\Im\frac{e^{\theta i} (\frac{2^N-e^{N\theta i}}{2^N})}{2-{e^{\theta i}}}$$
$$\Im\frac{e^{\theta i} ({2^N-e^{N\theta i}})}{2^{N+1}-{2^N e^{\theta i}}}$$
$$\Im\frac{e^{\theta i} ({2^N-e^{N\theta i}})}{2^{N}(2-{e^{\theta i}})}$$
$$\Im\frac{e^{\theta i} ({2^N-e^{N\theta i}})}{2^{N}(2-{e^{\theta i}})} \cdot \frac{2+e^{\theta i}}{2+e^{\theta i}}$$
I got :
$$\frac{2^{N+1} \sin \theta + 2^N \sin 2\theta - 2\sin (N+1)\theta- sin (N+2)\theta}{2^N(4-\sin 2\theta)}$$
I have no idea how to brings it to the desired result. They don't look very similar either. What is wrong , or what can I do next ?
I also noticed the denominator needs to be in terms of $\cos \theta$ so I also used $sin2\theta=2\sin\theta \cos\theta$ but then I don't know how to get rid of the $\sin\theta$ :
$$\frac{2^{N+1} \sin \theta + 2^N \sin 2\theta - 2\sin (N+1)\theta- sin (N+2)\theta}{2^N(4-2\sin\theta\cos \theta)}$$
As far as I can see you are correct down to $$\frac{e^{\theta i} ({2^N-e^{N\theta i}})}{2^{N}(2-{e^{\theta i}})}\ .$$ But then you multiplied top and bottom by $2+e^{\theta i}$, which is not the conjugate of $2-e^{\theta i}$.
Remember that the conjugate of $a+bi$ is $a-bi$ if $a$ and $b$ are real numbers - but $e^{\theta i}$ is not real.
In fact the conjugate of $e^{\theta i}$ is $e^{-\theta i}$. You can confirm this by writing it in terms of $\cos\theta$ and $\sin\theta$ and finding the conjugate in the usual way.
So, you should have multiplied top and bottom by $2-e^{-\theta i}$. Then in the denominator you get $$(2-e^{\theta i})(2-e^{-\theta i}) =4-2(e^{\theta i}+e^{-\theta i})+e^{\theta i}e^{-\theta i} =4-2(2\cos\theta)+1 =5-4\cos\theta\ .$$ Will leave you to multiply out the numerator and take the imaginary part.