Take a look at
http://mathworld.wolfram.com/TrigonometryAnglesPi10.html
There they use the half-angle formulas to evaluate cos(pi/10) in terms of cos(pi/5)
There is a nice discussion on evaluating cos(pi/5) at
https://socratic.org/questions/how-do-i-evaluate-cos-pi-5-without-using-a-calculator
prove dat Cos(18°) = 1/4√10+2√5
2 answers
Construct an isosceles triangle ABC
with angle A = 36°
AB = AC = 1
which leads to angle B = angle C = 72°
Let BC = x
Draw BD to hit AC at B, so that angle CBD = 36°
This makes BC = BD = x and
it makes triangle ABC similar to triangle BCD
Also triangle ABD is now isosceles, making AD = x , nice!
But AC = 1, so DC = 1-x
using triange ABC and triangle BCD
x/1 = (1-x)/x
x^2 = 1-x
x^2 + x - 1 = 0
x = (-1 ± √5)/2 , but x has to be positive
x = (√5 - 1)/2
using the cosine law on triangle ABC
[ (√5 - 1)/2 ]^2 = 1^2 + 1^2 - 2(1)(1)cos36°
(5 - 2√5 + 1)/4 = 2 - 2cos36
2cos36° = 2 - (3 - √5)/2
2cos36° = (4 - 3 + √5)/2
cos36° = (1 + √5)/4 <----- which just happens to be 1/2 the golden ratio!!!
recall cos2A = 1 - 2sin^2 A -----> cos36° = 1 - 2sin^2 18°
I will leave it up to you to to find sin 18° from that.
with angle A = 36°
AB = AC = 1
which leads to angle B = angle C = 72°
Let BC = x
Draw BD to hit AC at B, so that angle CBD = 36°
This makes BC = BD = x and
it makes triangle ABC similar to triangle BCD
Also triangle ABD is now isosceles, making AD = x , nice!
But AC = 1, so DC = 1-x
using triange ABC and triangle BCD
x/1 = (1-x)/x
x^2 = 1-x
x^2 + x - 1 = 0
x = (-1 ± √5)/2 , but x has to be positive
x = (√5 - 1)/2
using the cosine law on triangle ABC
[ (√5 - 1)/2 ]^2 = 1^2 + 1^2 - 2(1)(1)cos36°
(5 - 2√5 + 1)/4 = 2 - 2cos36
2cos36° = 2 - (3 - √5)/2
2cos36° = (4 - 3 + √5)/2
cos36° = (1 + √5)/4 <----- which just happens to be 1/2 the golden ratio!!!
recall cos2A = 1 - 2sin^2 A -----> cos36° = 1 - 2sin^2 18°
I will leave it up to you to to find sin 18° from that.