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By considering $\sum\limits_{n=1}^N z^{2n-1}$, where $z=e^{i\theta},$ show that $$ \sum\limits_{n=1}^N \cos{(2n-1)} \theta = \frac{\sin(2N\theta)}{2\sin\theta}, $$ where $\sin\theta\neq0$

I approached this question by starting with that the sum of real part of $z^{2n-1}$ is $\cos (2n-1)\theta$, and then I stated that the first term is $a= \cos \theta + i \sin \theta$, and $r= (\cos \theta + i \sin \theta)^2$, and therefore said it's a geometric sum, so I used the formula of a geometric series, and after simplification I removed the imaginary parts and I ended up with $( \sin (2n+1) \theta - \sin \theta )/ (-2 \sin \theta \cos \theta)$

I can't take it to the form they have given :/ can anyone please tell me where I went wrong , or if how to simplify this.

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Jun 1, 2014 at 15:25
  • $\begingroup$ did you try $\cos (2k \theta - \theta) = \cos (2k \theta)\cos \theta + \sin(2k \theta) \sin \theta$? Then you get two simple sums that have a closed-form expression $\endgroup$
    – Alex
    Jun 1, 2014 at 15:27
  • $\begingroup$ Thank you I will check this math jax tutorial....no I didn't :/ because it says to do this problem considering that $\endgroup$
    – M.S.E
    Jun 1, 2014 at 15:31
  • $\begingroup$ @Alex Considering z^2n-1 $\endgroup$
    – M.S.E
    Jun 1, 2014 at 15:32

1 Answer 1

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HINT:

Using Geometric Series, $$\sum_{n=1}^Nz^{2n-1}=\frac{z\{(z^2)^N-1\}}{z^2-1}=z^N\frac{z^N-z^{-N}}{z-z^{-1}}$$

Now using de Moivre's Theorem $z^N=\cos N\theta+\sin N\theta, z^N-z^{-N}=2\sin N\theta$

Alternatively,

$$2\sin\theta\cos(2n-1)\theta=\sin2(n)\theta-\sin2(n-1)\theta$$ which gives Telescoping Series

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