We get #cos 36^circ# mildly indirectly from the double and triple angle formula for cosine. It's pretty cool how it's done, and has a surprise ending.
We'll focus on #cos 72^circ#. The angle #theta=72^circ# satisfies
#cos(2 theta) = cos(3 theta).#
Let's solve that for #theta#, recalling #cos x=cos a# has solutions #x = pm a + 360^circ k.#
#2 theta = \pm 3 theta + 360^circ k#
#5 theta = 360 ^circ k # or #-theta = 360^circ k#
#theta = 72^circ k#
That includes the #360^circ k# so we can drop the "or" part.
I'm not writing a mystery here (despite the surprise ending) so I'll mention that #cos(2(72^circ)) = cos(144^circ)=-cos(36^circ)# is also a valid solution and we see how it's related to the question.
#cos(2 theta) = cos(3 theta)#
#2 cos ^2 theta -1 = 4 cos^3 theta - 3 cos theta #
Now let #x= cos theta#
#2 x ^2 -1 = 4 x^3 - 3x #
#4 x^3 - 2x^2 - 3x +1 = 0#
We know #x=cos(0 \times 72^circ)=1# is a solution so #(x-1)# is a factor:
#(x - 1) (4 x^2 + 2x - 1) = 0#
The quadratic has roots
#x = 1/4 (-1 \pm sqrt{5})#
The positive one must be #cos 72^circ # and the negative one #cos 144^circ#.
#cos 144^circ = 1/4 (-1 - sqrt{5})#
#cos 36^circ = cos(180^circ - 144^circ)= -cos 144^circ = 1/4(1 + sqrt{5})#
That's the answer. The surprise is it's half the Golden Ratio!
