# De Moivre's theorem:

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!

• The guy's name is De Moivre. Dec 2 '14 at 7:04
• Sorry for the typo Dec 2 '14 at 7:04
• That summation is meaningless Dec 2 '14 at 7:05
• It's an exercise in the book I'm reading Dec 2 '14 at 7:08
• Dec 2 '14 at 8:49

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 $$\sum_{k=0}^{N} \sin (k\theta).$$
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 $$sin(N+1)Theta/2$$