# Thread: Algebra, Problems For Fun (23)

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

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

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

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

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

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

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

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 $\displaystyle [\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]$ to be a power of 2).