TITLE: Show $1 + 2 \sum_{n=1}^N \cos n x = \frac{ \sin (N + 1/2) x }{\sin \frac{x}{2}}$ for $x \neq 0$
QUESTION [6 upvotes]: For $x \neq 0$, $$ 1 + 2 \sum_{n=1}^N \cos n x = \frac{ \sin (N + 1/2) x }{\sin \frac{x}{2}} $$

REPLY [12 votes]: Here is a well known trigonometric trick
$$
1+2\sum\limits_{n=1}^N\cos (nx)=
1+\frac{1}{\sin(x/2)}\sum\limits_{n=1}^N 2\cos (nx)\sin (x/2)=\\
1+\frac{1}{\sin (x/2)}\sum\limits_{n=1}^N(\sin (nx+x/2)-\sin (nx-x/2))=\\
1+\frac{1}{\sin (x/2)}(\sin (Nx+x/2)-\sin (x/2))=\\
1+\frac{\sin (Nx+x/2)}{\sin (x/2)}-1=
\frac{\sin (N+1/2)x}{\sin (x/2)}
$$
And this is a complex analysis approach
$$
1+2\sum\limits_{n=1}^N\cos(nx)=
e^{i0x}+\sum\limits_{n=1}^N(e^{inx}+e^{-inx})=
$$
$$
\sum\limits_{n=-N}^N e^{inx}=
\frac{e^{-iNx}(e^{i(2N+1)x}-1)}{e^{ix}-1}=
\frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1}=
$$
$$
\frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}=
\frac{2i\sin(N+1/2)x}{2i\sin(x/2)}=
\frac{\sin(N+1/2)x}{\sin(x/2)}
$$