We get cos 36^circcos36∘ 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^circcos72∘. The angle theta=72^circθ=72∘ satisfies
cos(2 theta) = cos(3 theta).cos(2θ)=cos(3θ).
Let's solve that for thetaθ, recalling cos x=cos acosx=cosa has solutions x = pm a + 360^circ k.x=±a+360∘k.
2 theta = \pm 3 theta + 360^circ k2θ=±3θ+360∘k
5 theta = 360 ^circ k 5θ=360∘k or -theta = 360^circ k−θ=360∘k
theta = 72^circ kθ=72∘k
That includes the 360^circ k360∘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)cos(2(72∘))=cos(144∘)=−cos(36∘) is also a valid solution and we see how it's related to the question.
cos(2 theta) = cos(3 theta)cos(2θ)=cos(3θ)
2 cos ^2 theta -1 = 4 cos^3 theta - 3 cos theta 2cos2θ−1=4cos3θ−3cosθ
Now let x= cos thetax=cosθ
2 x ^2 -1 = 4 x^3 - 3x 2x2−1=4x3−3x
4 x^3 - 2x^2 - 3x +1 = 04x3−2x2−3x+1=0
We know x=cos(0 \times 72^circ)=1x=cos(0×72∘)=1 is a solution so (x-1)(x−1) is a factor:
(x - 1) (4 x^2 + 2x - 1) = 0(x−1)(4x2+2x−1)=0
The quadratic has roots
x = 1/4 (-1 \pm sqrt{5})x=14(−1±√5)
The positive one must be cos 72^circ cos72∘ and the negative one cos 144^circcos144∘.
cos 144^circ = 1/4 (-1 - sqrt{5})cos144∘=14(−1−√5)
cos 36^circ = cos(180^circ - 144^circ)= -cos 144^circ = 1/4(1 + sqrt{5})cos36∘=cos(180∘−144∘)=−cos144∘=14(1+√5)
That's the answer. The surprise is it's half the Golden Ratio!
