I'm new to this site, english is not my mother tongue, and I'm just learning LaTeX. I'm basically a noob, so please be indulgent if I break any rule or habits.

I'm stuck at proving the following equation. I suppose I should use the formula for geometric series ($\sum\limits_{k=0}^{n}q^k=\frac{1-q^{n+1}}{1-q}$), and also use somewhere that $e^{î\theta}=cos(\theta)+i\sin(\theta)$

So here is the equation I have to prove : $$\frac{1}{2}+cos(\theta)+cos(2\theta)+...+cos(n\theta)=\frac{sin(n+\frac{1}{2}\theta)}{2sin(\frac{\theta}{2})}$$

Thanks for your help


marked as duplicate by lab bhattacharjee, graydad, muaddib, colormegone, saz Aug 1 '15 at 19:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $$ \dfrac{1}{2} + \sum_{r=1}^{n} \cos(r\theta) = \Re \left\{ \dfrac{1}{2} + \sum_{r=1}^{n} e^{ir\theta} \right\} = \Re \left\{ -\dfrac{1}{2} + \sum_{r=0}^{n} e^{ir\theta} \right\} $$ You should be able to use the formula for a geometric series on the final sum. $\endgroup$ – Kari Aug 1 '15 at 14:38
  • 2
    $\begingroup$ Incidentally you can also do this without using complex numbers by multiplying the left hand side by $2\sin(\frac{\theta}{2})$ and then expressing $2\sin A\cos B$as the difference of sines so that you get a telescoping series $\endgroup$ – David Quinn Aug 1 '15 at 15:04
  • $\begingroup$ That is one neat method, @DavidQuinn! $\endgroup$ – Kari Aug 1 '15 at 15:10
  • $\begingroup$ Related: math.stackexchange.com/questions/933402/… $\endgroup$ – colormegone Aug 1 '15 at 15:10
  • $\begingroup$ Thank you all. @RecklessReckoner, you found the related topic very quickly. You just remembered it or I missed some trick in the search tool? $\endgroup$ – Benji Aug 1 '15 at 15:55

$$ \begin{aligned} \dfrac{1}{2} + \sum_{r=1}^{n} \cos(r\theta) & = \Re \left\{ \dfrac{1}{2} + \sum_{r=1}^{n} e^{ir\theta} \right\} \\ & = \Re \left\{ -\dfrac{1}{2} + \sum_{r=0}^{n} e^{ir\theta} \right\} \\ & = \Re \left\{ -\dfrac{1}{2} + \dfrac{1-e^{(n+1)\theta}}{1-e^{i\theta}} \right\} \\ & = \Re \left\{ \dfrac{1-2e^{(n+1)\theta}+e^{i\theta}}{2(1-e^{i\theta})} \right\} \\ & = \Re \left\{ \dfrac{e^{-i\theta/2}-2e^{(n+1/2)\theta}+e^{i\theta/2}}{2(e^{-i\theta/2}-e^{i\theta/2})} \right\} \\ & = \Re \left\{ \dfrac{2\cos\left(\frac{\theta}{2}\right)-2\cos(n+1/2)\theta-2i\sin(n+1/2)\theta}{-2i\sin\left(\frac{\theta}{2}\right)} \right\} \end{aligned} $$

Multiply through by $1=\frac{i}{i}$, take the real part and there you have it. ^_^

  • $\begingroup$ Perfect, thanks. I think I gave up too quickly because I didn't think I could use the $\frac{i}{i}$ multiplication in the end to get rid of all the cos. $\endgroup$ – Benji Aug 1 '15 at 15:56

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