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What is the formula for: $$\sum_{i=1}^n \sin(i\theta)$$

I looked at various websites but cant find it. Does such a formula exist?

Tim Ratigan
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N.G.Tyson
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    I think you meant http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro, http://math.stackexchange.com/questions/297452/sines-and-cosines-of-angles-in-arithmetic-progression – lab bhattacharjee Jan 01 '14 at 06:50
  • Duplicate of http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – A.G. Jan 01 '14 at 08:28

2 Answers2

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Hint

Notice that $$\sin(k\theta)=\operatorname{Im}e^{ik\theta}$$ and use the geometric sum.

  • The first few terms are $1$, $e^{i\theta}$, $e^{i,2\theta}$, $e^{i,3\theta}$. Here $i = \sqrt{-1}$ not the index in your summation. You should write that as $\sum_{k=0}^{n} \sin(k\theta)$ – user44197 Jan 01 '14 at 08:36
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$\displaystyle\sum_{i=1}^{n}\sin (n\theta)$

=$Im(e^{in\theta})$

=$\displaystyle\sum_{i=1}^{n}Im(1+in\theta+\frac{(n\theta)^{2}}{2}-i\frac{(n\theta)^{3}}{6}+\frac{(n\theta)^{4}}{24}-i\frac{(n\theta)^{5}}{120}+...)$

$\displaystyle=\sum_{i=1}^{n}(n\theta-\frac{(n\theta)^{3}}{6}-\frac{(n\theta)^{5}}{120}-...)$

Kns
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