# Math Help - Algebra, Problems For Fun (23)

1. ## Algebra, Problems For Fun (23)

$[\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]=?$

2. $-\frac12\,=\,\cos120^\circ\,=\,cos(3\times40^\circ) \,=\,4\cos^340^\circ-3\cos40^\circ$

$\therefore\quad8\cos^340^\circ-6\cos40^\circ+1\ =\ 0$

$\implies\ \left(2\cos40^\circ\right)^3-3(2\cos40^\circ)+1\ =\ 0$

The polynomial $x^3-3x+1=0$ is irreducible over $\mathbb Z$ and hence over $\mathbb Q$ (by Gauß’s lemma); hence it is the minimum polynomial for $2\cos40^\circ$ over $\mathbb Q.$

$\therefore\ [\mathbb{Q}(2\cos 40^{\circ}):\mathbb{Q}]\,=\,[\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]\,=\,3$

The significance of this result is in showing that the angle 120° cannot be trisected using only compass and straightedge (since such a procedure would require $[\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]$ to be a power of 2).