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

Trigonometry

Fibonacci Numbers

  1. Ceva's Theorem: A Matter of Appreciation
  2. When the Counting Gets Tough, the Tough Count on Mathematics
  3. I. Sharygin's Problem of Criminal Ministers
  4. Single Pile Games
  5. Take-Away Games
  6. Number 8 Is Interesting
  7. Curry's Paradox
  8. A Problem in Checker-Jumping
  9. Fibonacci's Quickies
  10. Fibonacci Numbers in Equilateral Triangle
  11. Binet's Formula by Inducion
  12. Binet's Formula via Generating Functions
  13. Generating Functions from Recurrences
  14. Cassini's Identity
  15. Fibonacci Idendtities with Matrices
  16. GCD of Fibonacci Numbers
  17. Binet's Formula with Cosines
  18. Lame's Theorem - First Application of Fibonacci Numbers

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Copyright © 1996-2017 Alexander Bogomolny

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