$\cos(\pi/7)$ is a root of $8x^3-4x^2-4x+1=0$. How is the polynomial generated? According to Wolfram MathWorld, $\cos(\frac{\pi}{7})$ is a root of $8x^3-4x^2-4x+1=0$. Similarly, $\cos(\frac{2\pi}{7})$ is a root of $8x^3+4x^2-4x-1=0$. What's the procedure to generate these polynomials? I understand you can solve the cubic equations and check the cosines are indeed roots. My question is how does one arrive at those polynomials in the first place?

By DeMoivre's theorem,
$$(\cos x+i\sin x)^7=\cos7x+i\sin7x\quad\quad(*)$$
With $x=\frac\pi7$, the right hand side of $(*)$ reduces to $-1$. Expand the binomial on the left hand side and match up the real and imaginary parts (the latter of which is $0$), leaving us with
$$\cos^7x-21\cos^5x\sin^2x+35\cos^3x\sin^4x-7\cos x\sin^6x=-1$$
Rewrite each instance of $\sin^2x$ as $1-\cos^2x$ and simplify the result; you should end up with
$$64\cos^7x-112\cos^5x+56\cos^3x-7\cos x+1=0$$
the left hand side of which can be factorized as
$$(1+\cos x)(8\cos^3x-4\cos^2x-4\cos x+1)^2=0$$
but $\cos\frac\pi7\neq-1$,
$$8\cos^3\frac\pi7-4\cos^2\frac\pi7-4\cos\frac\pi7+1=0$$
Similarly, with $x=\frac{2\pi}7$, the right hand side of $(*)$ reduces to $1$, and so on.
