That's 40^circ, not one of the usual suspects. It's not constructible with straightedge and compass because we can't trisect anything but multiples of 90^circ.
Nonetheless cos({2pi}/9) is an algebraic number, the zero of a polynomial with integer coefficients.
cos(3 theta) = cos 120^circ
has roots
3 theta = \pm 120 ^circ + 360^circ k quad integer k
theta = \pm 40^circ + 120^circ k
cos(3 theta) = -1/2
4 cos^3 theta - 3 cos theta = -1/2
So cos 40^circ is one of the three roots of
4 x^3 - 3x = -1/2
That's called a depressed cubic (no x^2 term) and has an easy closed form solution. Rather than work it out, we can just get it from De Moivre:
cos theta + i sin theta = (cos 3 theta + i sin 3 theta)^{1/3}
cos theta - i sin theta = (cos 3 theta - i sin 3 theta)^{1/3}
cos theta = 1/2( (cos 3 theta + i sin 3 theta)^{1/3} + (cos 3 theta - i sin 3 theta)^{1/3})
cos theta = 1/2( (-1/2 + i sqrt{3}/2 )^{1/3} + (-1/2 - i sqrt{3}/2)^{1/3})
cos theta = 1/4( (-4 + 4 i sqrt{3})^{1/3} + (- 4 - 4 i sqrt{3})^{1/3})
The cube root of a complex number isn't that helpful. It's ambiguous as well.
