Prove $2\cos\frac{2\pi}{7}$ is root of $x^3 + x^2 - 2x -1$

Question

To prove that a regular septagon cannot be constructed by a straightedge and compass, it suffices to prove that $2\cos(\frac{2\pi}{7})$ is not constructible.

Several other answers to this question follow with the fact that $2\cos(\frac{2\pi}{7})$ is a root of

$$x^3 + x^2 - 2x -1$$

How do I derive that?

I know that since the cubic polynomial is irreducible in $\mathbb{Q}$, it has no real roots, implying that $\cos(\frac{2\pi}{7})$ is not constructible.

Answer 1

Let $a = \frac{2\pi}7$. Then, $ 4a = 2\pi-3a $ and

$$\cos4a = \cos3a$$

Use the identities $\cos 2t = 2\cos^2t -1$ and $\cos3t = 4\cos^3 t-3\cos t$ to expand the two sides

$$2(2\cos^2 a -1)^2 -1 = 4\cos^3a -3\cos a $$

After rearrangement

$$8\cos^4a -4\cos^3a -8\cos^2a +3\cos a+1=0$$

which factorizes as

$$(\cos a -1) (8\cos^3a + 4\cos^2a-4\cos a-1)=0$$

Since $\cos a \ne 1$, the equation above shows that $2\cos \frac{2\pi}7 $ satisfies

$$x^3+x^2 -2x-1=0$$

Answer 2

Because $$\left(2\cos\frac{2\pi}{7}\right)^3+\left(2\cos\frac{2\pi}{7}\right)^2-2\left(2\cos\frac{2\pi}{7}\right)-1=$$ $$=8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=$$ $$=8\cos^3\frac{2\pi}{7}-6\cos\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-2+2\cos\frac{2\pi}{7}+1=$$ $$=2\cos\frac{6\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{2\pi}{7}+1=$$ $$=\frac{2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}}+1=$$ $$=\frac{\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}}+1=0.$$

Answer 3

Let $z=e^{2\pi i/7}$, so $2\cos(2\pi/7)=z+\overline{z}=z+z^{-1}$. Plugging this into your cubic gives $$\begin{align*} (z+z^{-1})^3+(z+z^{-1})^2-2(z+z^{-1})-1&=z^3+3z+3z^{-1}+z^{-3}+z^2+2+z^{-2}-2z-2z^{-1}-1\\ &=z^3+z^2+z+1+z^{-1}+z^{-2}+z^{-3}.\end{align*}$$ This is a geometric series whose sum is $z^{-3}\frac{z^7-1}{z-1}$. But $z^7=1$, so this is $0$.

(This of course does not explain how you would come up with this seemingly magical polynomial $x^3+x^2-2x-1$. One way to do that is to work the argument above backwards. You start by knowing that $z$ is a root of $\frac{z^7-1}{z-1}=z^6+z^5+z^4+z^3+z^2+z+1$. To write this in terms of $z+z^{-1}$, you first multiply by $z^{-3}$ to get the more symmetric $z^3+z^2+z+1+z^{-1}+z^{-2}+z^{-3}=0$. Then you can group the terms $z^3+z^{-3}$ and $z^2+z^{-2}$, and figure out how to write each of these in terms of powers of $z+z^{-1}$, to write the whole thing as a polynomial in $z+z^{-1}$.

A slower but more systematic way would be to write the multiplication by $z+z^{-1}$ on the field $\mathbb{Q}(z)$ as a $6\times 6$ matrix using the basis $1,z,z^2,\dots,z^5$ (and the relation $z^6+z^5+\dots+1=0$) and then use linear algebra to find the minimal polynomial of this matrix.)