cos 36°

The purpose of this page is to establish that

It's a good exercise in trigonometry that was also useful in solving a curious sangaku problem.

We start with a regular pentagon. As every regular polygon, this one too is cyclic. So that we may assume its vertices lie on a circle and sides and diagonals form inscribed angles.

It follows that every angle of the regular pentagon equals 108°. Inscribed CAD is half of the central angle of 360°/5, i.e.

CAD = 36°.

By symmetry BAC = DAE implying that this two are also 36°. (Note in passing that we just proved that angle 108° is trisectable.) Other angles designated in the diagram are also easily calculated.

As far as linear segments are concerned, we may observe that triangles ABP, ABE, AEP are isosceles, in particular,

Triangles ABE and AEP are also similar, in particular

For the ratio x = AB/EP we have the equation

with one positive solution x = φ, the golden ratio. (This calculations were in fact performed to justify a construction of regular pentagon. We return to them here in order to facilitate the references.)

In ΔAEP AE = AB and EP is one of the sides such that AE/EP = φ. Drop a perpendicular from P to AE to obtain two right triangles. Then say

cos(AEP) = (AE/2)/EP = (AE/EP)/2 = φ/2.

But AEP = 36° and we get the desired result.

Using cos(36°) = (1 + √5)/4 we can find

cos(72°) = 2cos²(36°) - 1   = 2[(√5 + 1) / 4]² - 1   = (6 + 2√5) / 8 - 1   = (3 + √5) / 4 - 1   = (√5 - 1) / 4.

This is of course equal to sin(18°) as might have been expected from the general formula, sin(α) = cos(90° - α).

Also,

sin(72°) = √2(5 + √5 )/4.