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!
