De Moivre's theorem:

Question

Could someone help me to expand and express:

$$ \sum_{k=0}^N \cos(k\theta) $$

And:

$$ \sum_{k=0}^N \sin(k\theta) $$

In terms of $$\cos\theta/2$$ and $$\sin\theta/2$$

Using De Moivre's theorem:$$(\cos\theta + i \sin \theta)^N=\cos N\theta+i\sin N\theta$$

I'm still learning series and still not very good at them, so I need help, thanks!

Answer 1

You may write $$ \begin{align} \sum_{k=0}^{N} \cos (k\theta)&=\Re \sum_{k=0}^{N} e^{ik\theta}\\\\ &=\Re\left( \frac{e^{i(N+1)\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Re\left( \frac{e^{i(N+1)\theta/2}\left(e^{i(N+1)\theta/2}-e^{-i(N+1)\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Re\left( e^{iN\theta/2}\frac{\sin(N\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Re\left( \left(\cos (N\theta/2)+i\sin (N\theta/2)\right)\frac{\sin(N\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin(N\theta/2)}{\sin(\theta/2)}\cos (N\theta/2)\\\\ &=\frac{\sin(N\theta)}{2\sin(\theta/2)} \end{align} $$ You easily obtain $$ \sum_{k=0}^{N} \sin (k\theta). $$

Answer 2

In my opinion there is a mistake in the solution, being $$(exp(i(N+1)Theta/2)-exp(-i(N+1)Theta/2))/2i$$ equal to $$sin(N+1)Theta/2$$