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.
Best Answer
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