Proving $1+\cos a + \cos 2 a+ \cdots + \cos(n-1)a = 0$, when $a=2\pi/n$ and $n$ is odd

Question

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 $n$.

Answer 1

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

Answer 2

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

Answer 3

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.

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

Answer 4

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

Answer 5

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

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