Proving $1+\cos a + \cos 2 a+ \cdots + \cos(n-1)a = 0$, when $a=2\pi/n$ and $n$ is odd I am trying to prove that:
$1+\cos a+\cos 2a+\cos3a+\cos4a=0$
where $a=\frac{2\pi}5$ (pentagon arrangement).
Actually this is true for any $n>1$:
$1+\cos a+\cos2a+\dots+\cos(n-1)a=0$ (polygon) where $a={2 \pi\over n}$.
Easy to show for even $n$ since the $\cos$ cancel themselves 2 at a time
but but for odd $n$ (say 5)?
This comes from the fact that if you have $n$ same objects equally space around a unit circle, the center of gravity has to be at the origin, so sum of sines equals zero (easy) and sum of cosine also, not so easy for odd $n$.
 A: If you are familiar with complex number:
\begin{align}
\sum_{k=0}^{n-1} \cos\left( \frac{2k\pi}n\right)&= \Re \left(\sum_{k=0}^{n-1} \exp\left( \frac{2ik\pi}n\right)\right) \\
&=\Re\left(\frac{1-\exp\left(\frac{2in\pi}{n} \right)}{1-\exp\left(\frac{2i\pi}{n} \right)} \right)\\
&=\Re\left(\frac{1-\exp\left(2i\pi \right)}{1-\exp\left(\frac{2i\pi}{n} \right)} \right)\\
&=0
\end{align}
since $\exp(2\pi i)=1$.
A: I will prove the general identity here. Note that $e^{2\pi i/n}=e^{ia}$ is a root of $x^n-1=(x-1)(x^n+x^{n-1}+\dots+1)=0$. Since $e^{ia}\ne1$, we have
$$\sum_{k=0}^{n-1}e^{kia}=0$$
Using Euler's identity $e^{ix}=\cos x+i\sin x$ to extract the real part of this equation gives the desired result:
$$\sum_{k=0}^{n-1}\cos ka=0$$
A: If you want to prove using vectors , then,
Let $\overrightarrow { v_{n}}$ be a vector which represents the $n^{th}$ side of a $n$ sided polygon,
Again  $\overrightarrow {v_{n}}$ can be written as $a_{n} \widehat{i} + b_{n} \widehat {j}$
Now by polygon law of vector addition , we get,
$\sum_{0}^{n} \overrightarrow {v_{n}} =0$
Therefore $\sum_{0}^{n} a_{n} \widehat{i} + b_{n} \widehat{j} =0$
Hence, $\sum_{0}^{n} a_{n} =0$
$a_{n} =\mid \overrightarrow {v_{n}}\mid \cdot \cos \alpha_{n}$ , where $\alpha_{n}$ is the angle made by the $n^{th}$ side of the polygon with the $x$-axis.
$\therefore \sum_{0}^{n} \cos \alpha_{n} =0$
A: From the angle sum and difference formulas
$$2\sin\frac12a\cos na=\sin\left(n+\frac12\right)a-\sin\left(n-\frac12\right)a$$
Then
$$\begin{align}2\sin\frac12a\sum_{k=0}^{n-1}\cos ka&=\sum_{k=0}^{n-1}\left[\sin\left(k+\frac12\right)a-\sin\left(k-\frac12\right)a\right]\\
&=\sin\left(n-\frac12\right)a+\sin\frac12a\end{align}$$
In this case $a=\frac{2\pi}n$ so
$$\sin\left(n-\frac12\right)a=\sin\left(2\pi-\frac12a\right)=-\sin\frac12a\ne0$$
So
$$\sum_{k=0}^{n-1}\cos ka=0$$
A: It follows from radial symmetry of physical forces acting on a particle by end to end positioning and adding vectors forming a closed regular pentagon/polygon. Also it might occur as more general or natural if we ponder a bit on how trigonometrical ratio definitions came into being.
It holds good even if start angle of first vector to $x$ axis is non-zero.
If $n$ forces each of magnitude $F$ act on a point start first  from $x-$ axis then by statics force equilibrium projections
On $x$ axis
$$ F( 1+\cos a+\cos2a+\dots+\cos(n-1)a ) =0 $$
On $y$ axis
$$ F( 0+\sin a+\sin 2a+\dots+\sin(n-1)a ) =0. $$
