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I have not been able to completely solve this problem and it's driving me crazy. Could you please help.
The question is to show that, $$\sum_{n=1}^N \frac{\sin n\theta}{2^n} =\frac{2^{N+1}\sin\theta+\sin N\theta-2\sin(N+1)\theta }{2^N(5-4\cos\theta)}$$ Where do I start? I tried solving this using de Moivre's Theorem but I don't know where I am going wrong. Could you please help me or if possible show other ways to tackle this particular problem.

Any Help is much appreciated!

Thanks in Advance!

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Well your summation is the imaginary part of $\sum_{n = 1}^{N}\frac{e^{in\theta}}{2^{n}} = \sum_{n = 1}^{N}(e^{i\theta}/2)^{n}$. Then it is just a finite summation of a geometric series. – ADF Mar 26 '12 at 4:06
Please edit your question as the displayed equation is seriously messed up. – Gerry Myerson Mar 26 '12 at 5:09
@GerryMyerson Oh, sorry about that. Probably forgot to press the Shift key along with the = sign :) – funktor Mar 26 '12 at 5:34
up vote 6 down vote accepted

If you follow one of the suggestions the summation is the imaginary part of

$$ \begin{align*} \sum_{n = 1}^{N}\frac{e^{in\theta}}{2^{n}} &= \sum_{n = 1}^{N}(e^{i\theta}/2)^{n}\\ &= \frac{e^{i\theta}}{2} \frac{\left(1-\frac{e^{Ni\theta}}{2^N}\right)}{\left(1-\frac{e^{i\theta}}{2}\right)}\\ &= \frac{e^{i\theta}(2^N-e^{Ni\theta})}{2^N(2-e^{i\theta})}\\ &= \frac{e^{i\theta}(2^N-e^{Ni\theta})(2-e^{-i\theta})}{2^N(2-e^{i\theta})(2-e^{-i\theta})}\\ &= \frac{(2^Ne^{i\theta}-e^{(N+1)i\theta})(2-e^{-i\theta})}{2^N(4-2(e^{i\theta}+e^{-i\theta})+1)}\\ &= \frac{2^{(N+1)}e^{i\theta}-2e^{(N+1)i\theta}-2^N+e^{Ni\theta}}{2^N(4-2(e^{i\theta}+e^{-i\theta})+1)} \end{align*} $$

The imaginary part of this is

$$ \frac{2^{(N+1)}\sin \theta - 2\sin (N+1)\theta + \sin N\theta}{2^N (5-4\cos \theta)}$$

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Hint: Write $\sin(n\theta)=\dfrac{e^{in\theta}-e^{-in\theta}}{2i}$, then, use the formula for the sum of a geometric series.

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@ADF's suggestion nearly halves the task, don't you think? – Did Mar 26 '12 at 6:03
@Dider: After computing the first sum, one would probably realize that the second sum is the conjugate of the first, and the work is cut in half. – robjohn Mar 26 '12 at 10:30
Your suggestion (in the comment, not in the post) amounts to using the fact that the sine is the imaginary part of the complex exponential. I agree; thus it would be more logical to use this fact and to base the proof on imaginary parts from its onset, wouldn't it? – Did Mar 26 '12 at 10:49

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