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.
