Draw
a 36^{o}-72^{o}-72^{o} triangle, with two long sides =
1, and the short side = x .

Now, bisect
one of the 72° angles forming two triangles:

a 36^{o}-36^{o}-108^{o
}isosceles triangle whose short sides are both x , and a smaller 36^{o}-72^{o}-72^{o}
isosceles triangle whose long sides are x , and short side is 1−x

Since
the two 36^{o}-72^{o}-72^{o} triangles are
proportional: <o:p></o:p>

x/1=(1-x)/x

solving x^{2}+x-1=0

and
ignoring the negative answer

we
get x=(-1+sqrt(5))/2

a.k.a.
phi, the golden ratio<o:p></o:p>

Now
bisect the obtuse angle in the 36^{o}-36^{o}-108^{o}
triangle to get a right

(36^{o}-54^{o}-90^{o}) triangle, with a hypotenuse=(-1+sqrt(5))/2 and the long leg =1/2 .

so:

**cos 36°****
=1/(-1+sqrt(5))=(1+phi)/2**** =.809**
approximately.