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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 $n$.

Blue
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user539511
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    Look at the sum of the roots of $x^5+0x^4-1=0$ and in particular the sum of their real parts. No computation or sum of geometric series necessary. – YAlexandrov Mar 11 '18 at 04:29
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    I think the easiest way to see this would be to work with complex exponentials, where we get a geometric sum. This also deals with the real and imaginary parts at the same time. – Tob Ernack Mar 11 '18 at 04:30
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    This comes from the fact ... That is good geometric insight into it, and does in fact easily extend to the odd case. Consider that a regular polygon with an odd number of vertices $n$ has $n$ different axes of symmetry (the diameters through each vertex). By symmetry, the center of gravity must lie on each and every axis of symmetry, and therefore on their intersection, which is the center of the circle. – dxiv Mar 11 '18 at 04:42

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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 $\exp(2\pi i)=1$.

Siong Thye Goh
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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: $$\sum_{k=0}^{n-1}\cos ka=0$$

Parcly Taxel
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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.

$\therefore \sum_{0}^{n} \cos \alpha_{n} =0$

Vishaal Selvaraj
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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 $$\sum_{k=0}^{n-1}\cos ka=0$$

user5713492
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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

$$ F( 0+\sin a+\sin 2a+\dots+\sin(n-1)a ) =0. $$

Narasimham
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