# How to express $\cos(20^\circ)$ with radicals of rational numbers?

In showing that the trisection of an angle with ruler and compass is not possible in general one shows that $\cos(20^\circ)$ cannot be constructed (thus the angle $60^\circ$ cannot be trisected) by determining its minimal polynomial, which is $8x^3-6x-1$.

Solving $8x^3-6x-1=0$ yields a solution $x_1=\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}+\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$. Expressing $\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}$ and $\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$ in polar form yields $x_1=\cos(20^\circ)$.

Is it possible to express $\cos(20^\circ)$ with radicals without complex numbers?