cos 36°

The purpose of this page is to establish that

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

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 \(72^{\circ}=\frac{360^{\circ}}{5}\), 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,

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

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

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

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

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

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

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

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

which is immediately verified by squaring the two expressions. Also

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

from \(\mbox{sin}2\alpha = 2\space \mbox{sin}\alpha \space \mbox{cos}\alpha\).

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

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

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

And, of course,

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

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

Trigonometry

Fibonacci Numbers

Golden Ratio

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