Sum up trigonometric series $$\cos \frac{2π}{2013} +\cos \frac{4π}{2013} +\cdots+\cos \frac{2010π}{2013} + \cos \frac{2012π}{2013}$$
How to sum it up?
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Original Approach: Using the formula for the sum of a geometric series, we get
$$
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}=\frac{e^{2\pi i}-e^0}{e^{\frac{2\pi i}n}-1}=0\tag{1}
$$
and using Euler's Formula
$$
\begin{align}
\sum_{k=0}^{n-1}e^{\frac{2\pi i}nk}
&=\sum_{k=0}^{n-1}\cos\left(\frac{2\pi}nk\right)+i\sum_{k=0}^{n-1}\sin\left(\frac{2\pi}nk\right)\tag{2}
\end{align}
$$
Substituting $n\mapsto2n+1$ and subtracting the $k=0$ term yields
$$
\sum_{k=1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)=-1\tag{3}
$$
Substituting $k\mapsto2n+1-k$ shows that
$$
\sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right)
=\sum_{k=n+1}^{2n}\cos\left(\frac{2\pi}{2n+1}k\right)\tag{4}
$$
Combine $(3)$ and $(4)$ to get the desired result.

Alternate Approach: Using the formula for the sum of a geometric series, we get
$$
\begin{align}
\sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k}
&=\frac{e^{\frac{2\pi i}{2n+1}(n+1)}-e^{\frac{2\pi i}{2n+1}}}{e^{\frac{2\pi i}{2n+1}}-1}\\
&=\frac{e^{\pi i}-e^{\frac{\pi i}{2n+1}}}{e^{\frac{\pi i}{2n+1}}-e^{-\frac{\pi i}{2n+1}}}\\
&=\frac{-1-\cos\left(\frac\pi{2n+1}\right)-i\sin\left(\frac\pi{2n+1}\right)}{2i\sin\left(\frac\pi{2n+1}\right)}\\
&=\frac i2\cot\left(\frac\pi{4n+2}\right)\color{#C00000}{-\frac12}\tag{5}
\end{align}
$$
and using Euler's Formula
$$
\begin{align}
\sum_{k=1}^ne^{\frac{2\pi i}{2n+1}k}
&=\color{#C00000}{\sum_{k=1}^n\cos\left(\frac{2\pi}{2n+1}k\right)}+i\sum_{k=1}^n\sin\left(\frac{2\pi}{2n+1}k\right)\tag{6}
\end{align}
$$
The real parts of $(5)$ and $(6)$ give the desired result.
