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?
*Calculator is not allowed.
 A: 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.
A: okay Ill give you the hint Note the angles are in AP with common difference $d=2\pi/2013$ and $a=2\pi/2013$   so general formula for such series is $\frac{cos(a+\frac{(n-1)d}{2})}{sin\frac{d}{2}}.sin\frac{nd}{2}$ here $a$= first angle $d$=common difference between two angles and $n$= number of terms. Hope it helps you.
A: Let $\alpha=\frac{\pi}{2013}$ and let
$$S=cos2\alpha+cos4\alpha+cos6\alpha+\cdots+cos2010\alpha+cos2012\alpha$$ Multiply both sides with $2sin\alpha$ and use
$$2sinAcosB=sin(A+B)+sin(A-B)$$ so we get
$$(2sin\alpha) \:S=2sin\alpha cos2\alpha+2sin\alpha cos4\alpha+2sin\alpha cos6\alpha+\cdots+2sin\alpha cos2010\alpha+2sin\alpha cos2012\alpha$$ $\implies$
$$(2sin\alpha)S=sin3\alpha-sin\alpha+sin5\alpha-sin3\alpha+sin7\alpha-sin5\alpha+\cdots+sin2011\alpha-sin2009\alpha+sin2013\alpha-sin2011\alpha$$ $\implies$
$$(2sin\alpha)S=-sin\alpha$$
$$S=\frac{-1}{2}$$
