<s>[INST] 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?
 [/INST] $$
\cos (20^\circ) = \cos (\pi/9) = -\frac12 (-1)^{8/9} \left(1+(-1)^{2/9}\right)
$$
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