# cos 36°

The purpose of this page is to establish that

\(\cos (36^{\circ}) = \displaystyle\frac{(1 + \sqrt{5})}{4}\).

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 \(108^{\circ}\). Inscribed \(\angle CAD\) is half of the central angle \(72^{\circ}=\displaystyle\frac{360^{\circ}}{5}\), i.e.

\(\angle CAD = 36^{\circ}\).

By symmetry \(\angle BAC = \angle DAE\) implying that these two are also \(36^{\circ}\). (Note in passing that we just proved that angle \(108^{\circ}\) 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,

\(AB = BP\),

\(AB = AE\),

\(AP = EP\).

Triangles ABE and AEP are also similar, in particular

\(BE / AB = AE / EP\), or

\(BE\times EP = AB^{2}\), i.e.

\((BP + EP)\times EP = AB^{2}\), and lastly,

\((AB + EP)\times EP = AB^{2}\).

For the ratio \(x = AB/EP\) we have the equation

\(x + 1 = x^{2}\),

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

In \(\triangle AEP\), \(AE = AB\) and \(EP\) is one of the sides such that \(AE/EP = \phi \). Drop a perpendicular from \(P\) to \(AE\) to obtain two right triangles. Then say,

\(\cos (\angle AEP) = (AE/2)/EP = (AE/EP)/2 = \phi /2\).

But \(\angle AEP = 36^{\circ}\) and we get the desired result.

Using \(\cos (36^{\circ}) = (1 + \sqrt{5})/4\) we can find

\(\cos (18^{\circ})=\sqrt{2(5+\sqrt{5})}/4\)

from \(\cos 2\alpha = 2\cos ^{2}\alpha - 1\) and then

\(\sin (18^{\circ})=\sqrt{2(3-\sqrt{5})}/4\)

from \(\cos ^{2}\alpha + \sin ^{2}\alpha = 1\). Now, it may be hard to believe but this expression simplifies to

\(\sin (18^{\circ}) = (\sqrt{5} - 1)/4\),

which is immediately verified by squaring the two expressions. Also

\(\sin (36^{\circ})=\sqrt{2(5-\sqrt{5})}/4\)

from \(\sin 2\alpha = 2\space \sin \alpha \space \cos \alpha\).

We can easily find \(\cos (72^{\circ})\) from \(\cos 2\alpha = 2\cos ^{2}\alpha - 1\):

\( \begin{align} \cos (72^{\circ}) &= 2\cos ^{2}(36^{\circ}) - 1 \\ &= 2[(\sqrt{5} + 1) / 4]^{2} - 1 \\ &= (6 + 2\sqrt{5} / 8 - 1 \\ &= (3 + \sqrt{5} / 4 - 1 \\ &= (\sqrt{5} - 1) / 4. \end{align} \)

This is of course equal to \(\sin (18^{\circ})\) as might have been expected from the general formula, \(\sin \alpha = \cos (90^{\circ} - \alpha )\).

And, of course,

\(\sin (72^{\circ})=\sqrt{2(5+\sqrt{5})}/4\)

because \(\sin (72^{\circ}) = \cos (18^{\circ})\).

Finally, let's compute $\sin 54^{\circ}:\;$

$\displaystyle\begin{align} \sin 54^{\circ} &= \sin 36^{\circ}\cos 18^{\circ}+\cos 36^{\circ}\sin 18^{\circ}\\ &= \displaystyle\frac{\sqrt{2(5-\sqrt{5})}}{4}\cdot \displaystyle\frac{\sqrt{2(5+\sqrt{5})}}{4} + \displaystyle\frac{1 + \sqrt{5}}{4}\cdot\displaystyle\frac{\sqrt{5}-1}{4}\\ &=\displaystyle\frac{\sqrt{5}+1}{4}, \end{align}$

as expected since $\sin 54^{\circ}=\cos 36^{\circ}.$

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