I tried the following approach.

Let \( z = \dfrac{ e^{ i \theta }}{2} \). Then using de Moivre's theorem:

\[ \sum _ { n = 1 } ^ { N } \frac{ \sin n \theta } { 2 ^ n } = \text { Im } \left ( \sum _ { n = 1 } ^ { N } z^ { n} \right) \]

This is sum of GP. Using the sum of GP:

\[ \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ \frac{e ^{ i \theta } } { 2 } ( 1 - \frac{e ^{ N i \theta } } {2^ { N } } ) } { 1- \frac{ e^{i \theta } } { 2 } }\]

This simplifies to:

\[ \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ e ^{ i \theta } ( 2 ^ { N } - e ^{ N i \theta } ) } { 2 ^ { N} ( 2- e^{i \theta } ) }\]

\[ \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ ( 2 ^ { N } e ^{ i \theta } - e ^{ (N + 1 ) i \theta } ) } { 2 ^ { N} ( 2- e^{i \theta } ) } ~~~~~ (*)\]

Now I changed the complex numbers on the RHS from Euler's form to cis form, then made the denominator real and finally took imaginary part of the expression. Is there a simpler way to solve this problem? My method is very complicated, thus error-prone.

In fact, I have already made an error which is why I have posted this note. I am getting the RHS as

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

i.e. without the \(\sin N \theta \) term in the numerator. Where is the \( \sin N \theta \) term coming from? There are only \(e^{ i \theta } \) and \( e ^{ (N + 1 ) i \theta }\) in the \((*)\) equation. I can't seem to find out where I have gone wrong.

Please help.

Thanks.

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## Comments

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TopNewesttake \(sin(x)=\dfrac{-i}{2}(e^{ix}-e^{-ix}\). show this is true by de moivres.

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Thanks for helping Aareyan, this method is nice! :)

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I think you are missing out any term. I solved the RHS and got the desired result. Recheck it. As far as i think this is a proofing problem and the method you followed is very simple and short i think. Just need some care while solving.

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Thanks Sachin :), I realized I was silly mistakes. I solved it again carefully and now I got the desired result.

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