# An Algebraic approach to determining cos(2pi/5)

The identity

$\cos \left ( \cfrac{2\pi}{5}\right ) = \cfrac{1}{4}(\sqrt{5}-1)$

can be determined most directly by studying the pentagram and arriving at the value by observing certain similarity relations between different lengths within the pentagram. That approach however requires many preliminaries in geometry and specific knowledge about pentagrams, which is great if you already know them, but which makes it cumbersome if all you want is to dispel some doubts as to whether its true. I’d therefore would like to lay out an alternate approach which requires only fairly rudimentary knowledge about polynomials and complex numbers, though still quite a lot. Okay. Let’s make the Faustian bargain.

The approach here is to solve the equation

$z^5 - 1 = 0$

since this equation has five roots of the form $\alpha_ k= \cos(2\pi k / 5) + i\sin(2\pi k / 5)$ so $\cos(2\pi / 5)$ will be the real part of one of the solutions.

To solve the equation we first factor out $z - 1$ using either polynomial division or the generalized conjugate rule.

$z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)$

Plan: To find the real part of the roots $\alpha$ of the equation $z^4 + z^3 + z^2 + z + 1 = 0$

Firstly let us remark:

The equation $z^4 + z^3 + z^2 + z + 1 = 0$ has no real solutions and since the polynomial has real coefficients the 4 non-real solutions come in conjugate pairs.

Now the conjugate pair part is important since this means that if $\alpha$ is a root then so is $\bar{\alpha}$ and the polynomial  $z^4 + z^3 + z^2 + z + 1$ must contain the factor

$(z - \alpha)(z - \bar{\alpha}) = z^2 - (\alpha + \bar{\alpha})z + |\alpha|^2$

Since the roots $\alpha$ all lie on the unit circle we have $|\alpha|= 1$ and further we note that $\alpha + \bar{\alpha}$ is twice the real part of $\alpha$ which is precisely the quantity we want to determine. So what we really need to find is the coefficient $a$  and $b$ in the factorization

$z^4 + z^3 + z^2 + z + 1 = (z^2- az + 1)(z^2 - bz + 1)$.

Computing the product we get

$(z^2- az + 1)(z^2 - bz + 1) = z^4 - (a + b)z^3 + (2 + ab)z^2 - (a + b)z + 1$

And then by comparing coefficients we get the system of equations

$\begin{cases}a + b = -1 & \\ ab = -1\end{cases}$

Oh look! It’s a good old sum-product system of equations which can be solved by solving the quadratic equation

$x^2 - (a + b)x - ab = 0$

$x^2 + x - 1 = 0$

which has the solutions

$x = \cfrac{1}{2}(-1 \pm \sqrt{5})$

or

$\begin{cases}a = \cfrac{1}{2}(\sqrt{5} - 1)& \\ b = -\cfrac{1}{2}(\sqrt{5} + 1) &\end{cases}$

And finally

$\cos(2\pi/5) = \cfrac{a}{2} = \cfrac{1}{4}(\sqrt{5} - 1)$

since this is largest of the two roots.