How do you evaluate #Cos((2pi)/9)#?

Question

Dylan Arnold · Accepted Answer

It's one of the roots of #4 x^3 - 3x = -1/2 # and has a closed form involving complex cube roots:

# cos ({2pi}/9) = 1/4( (-4 + 4 i sqrt{3})^{1/3} + (- 4 - 4 i sqrt{3})^{1/3}) quad # near #.766# 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.