psycholeo
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Excuse if my Latex doesn't work

Prove

\sum_{r=1}^n cos(r\theta)$$ = $$\frac{cos(0.5(n+1)\theta)(sin(  0.5n\theta)}{sin(0.5\theta)}
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Adacic
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Just copy and paste
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psycholeo
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(Original post by Adacic)
Just copy and paste
I did I copied and pasted it from LaTeX and it didn't work
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Adacic
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Σ d
Idk
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ghostwalker
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(Original post by psycholeo)
Excuse if my Latex doesn't work



$$\sum_{r=1}^n cos(r\theta)$$ = $$\frac{cos(0.5(n+1)\theta)(sin( 0.5n\theta)}{sin(0.5\theta)}$$
Use [t e x] and [/ t e x] without the spaces around the whole thing and ditch the dollar signs.
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psycholeo
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(Original post by ghostwalker)
Use [t e x] and [/ t e x] without the spaces around the whole thing and ditch the dollar signs.
Thanks
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ghostwalker
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(Original post by psycholeo)
Thanks
np.

Additionally, if you make your first command "\displaystyle" the typset is slightly larger, and the limits on the summation will be above and below the sigma, rather than after it.

Anyhow, to the question itself.

Have you made any progress/working?
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psycholeo
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(Original post by ghostwalker)
np.

Additionally, if you make your first command "\displaystyle" the typset is slightly larger, and the limits on the summation will be above and below the sigma, rather than after it.

Anyhow, to the question itself.

Have you made any progress/working?
No i'm unsure of where to start in all honesty
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3pointonefour
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Try considering the sum of [cos(Rx) + isin(Rx)] from 1 to n, using exponentials.
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RDKGames
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(Original post by psycholeo)
Excuse if my Latex doesn't work

Prove

\sum_{r=1}^n cos(r\theta)$$ = $$\frac{cos(0.5(n+1)\theta)(sin(  0.5n\theta)}{sin(0.5\theta)}
It's easy to see that you are essentially summing up the real parts of complex numbers numbers with magnitude 1.

So if you denote \cos \theta + \mathbf{i} \cos \theta = e^{\mathbf{i} \theta}, and also note De Moivre's Theorem, then we have that:

$\begin{align*} \displaystyle \sum_{r=1}^n \cos (r \theta)  = \sum_{r=1}^n \Re (\cos r \theta + \mathbf{i} \sin r \theta) & = \Re \left( \sum_{r=1}^n \cos r\theta + \mathbf{i} \sin r\theta \right) \\ & = \Re \left( \sum_{r=1}^n (\cos  \theta + \mathbf{i} \sin \theta)^r \right) \\ & = \Re \left( \sum_{r=1}^n (e^{\mathbr{i}\theta})^r \right) \end{align*}$

Can you notice what sort of series this is, hence finish it off from there?

N.B. \Re(z) denotes \mathrm{Re}(z) which is the real part of the complex number z.
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psycholeo
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How would I do that? Sorry am really confused
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RDKGames
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(Original post by psycholeo)
How would I do that? Sorry am really confused
Which part confuses you?
Do you understand what I’ve written?
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3pointonefour
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(Original post by RDKGames)
It's easy to see that you are essentially summing up the real parts of complex numbers numbers with magnitude 1.

So if you denote \cos \theta + \mathbf{i} \cos \theta = e^{\mathbf{i} \theta}, and also note De Moivre's Theorem, then we have that:

$\begin{align*} \displaystyle \sum_{r=1}^n \cos (r \theta)  = \sum_{r=1}^n \Re (\cos r \theta + \mathbf{i} \sin r \theta) & = \Re \left( \sum_{r=1}^n \cos r\theta + \mathbf{i} \sin r\theta \right) \\ & = \Re \left( \sum_{r=1}^n (\cos  \theta + \mathbf{i} \sin \theta)^r \right) \\ & = \Re \left( \sum_{r=1}^n (e^{\mathbr{i}\theta})^r \right) \end{align*}$

Can you notice what sort of series this is, hence finish it off from there?

N.B. \Re(z) denotes \mathrm{Re}(z) which is the real part of the complex number z.
Never seen that R notation before
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DFranklin
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(Original post by 3pointonefour)
Never seen that R notation before
It's somewhat a TeX thing (i.e. that notation is what you get using \Re in TeX) and I have to admit I don't particularly like it myself.

https://tex.stackexchange.com/questi...maginary-parts
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RDKGames
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(Original post by 3pointonefour)
Never seen that R notation before
In addition to what DFranklin said, I also admit that I don't particularly like it, and simply used it here because I'm lazy to put \mathrm{Re} everywhere as opposed to just \Re when typesetting.
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username4240292
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(Original post by 3pointonefour)
Never seen that R notation before
If you mean you aren't familiar with taking real/imaginary parts or complex numbers in general, there's a way to get it without using them, but it's probably easier with them.
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3pointonefour
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(Original post by ftfy)
If you mean you aren't familiar with taking real/imaginary parts or complex number in general, there's a way to get it without using them, but it's probably easier with them.
I meant I've never seen that fancy R notation used to denote Re(z). But I know what Re(z) and Im(z) is.
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