possibility of trisection of angles

Question

i know that $\frac{\pi}{7}$ can be trisected if and only if $4x^3-3x-cos( \frac{\pi}{7})$ is reducible over $\mathbb {Q}$$(cos\frac{\pi}{7}) $.

but i don't know how to check this.

help pls.

thanks

Answer 1

By playing a bit with the cyclotomic polynomial $\Phi_n(x)$, we have that $\cos\frac{\pi}{n}$ is an algebraic number of degree $\frac{\phi(n)}{2}$ over $\mathbb{Q}$. This gives that $\cos\frac{\pi}{21}$ has degree $6$ and $\cos\frac{\pi}{7}$ has degree three over $\mathbb{Q}$. The roots of $$T_3(x)-\cos\frac{\pi}{7}$$ are $\beta=\cos\frac{\pi}{21},\cos\left(\frac{\pi}{21}+\frac{2\pi}{3}\right)=\frac{-\beta+\sqrt{3}\sqrt{1-\beta^2}}{2},\cos\left(\frac{\pi}{21}+\frac{4\pi}{3}\right)=\frac{-\beta-\sqrt{3}\sqrt{1-\beta^2}}{2}$, or $\cos\frac{\pi}{21},\cos\frac{5\pi}{7},-\cos\frac{8\pi}{21}$. Since $\cos\frac{5\pi}{7}$ belongs to $\mathbb{Q}\left(\cos\frac{\pi}{7}\right)$ due to $$\cos\frac{5\pi}{7}=T_5\left(\cos\frac{\pi}{7}\right),$$ in $\mathbb{Q}\left(\cos\frac{\pi}{7}\right)$ we can find $\beta$ and $T_8(\beta)$ by solving a quadratic equation.