Theorem of the convergence of the series of fourier!

Question

During the demonstration of the theorem of the convergence of the series of fourier, my teacher wrote :$$ \frac{1}{2}+ \sum_{k=1}^{n} \cos(ky)=\frac{\sin((n+\frac{1}{2})y)}{2\sin(\frac{y}{2})} $$

he doesn't explain how to get this for the lack of time, I'm curious so I tried alone, without success. How I can get this equality?

Answer 1

Very briefly: Write $$\cos(ky)=\frac{e^{iky}+e^{-iky}}{2}$$ and use the formula for the sum of a finite geometric series.

Edit: To amplify on that a little bit, you can now rearrange the sum on the left to $$\frac12\sum_{k=-n}^n e^{iky}.$$

Edit 2: To sum the above, write it as $$\frac12e^{-iny}\color{blue}{\sum_{j=0}^{2n} e^{ijy}} =\frac12e^{-iny}\color{blue}{\frac{1-e^{i(2n+1)y}}{1-e^{ijy}}} \cdot\color{green}{\frac{e^{-iy/2}}{e^{-iy/2}}} =\frac12\frac{e^{-i(n+1/2)y}-e^{i(n+1/2)y}}{e^{-iy/2}-e^{iy/2}} =\frac12\frac{\sin((n+1/2)y)}{\sin(y/2)}.$$

Answer 2

Multiply both sides by $2\sin(\dfrac{y}{2})$ and remark that

$2\sin(\dfrac{y}{2})\cos(ky) = \sin(\dfrac{2k+1}{2}y) - \sin(\dfrac{2k-1}{2}y)$