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cos 36°

The purpose of this page is to establish that

  cos(36°) = (1 + 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°. 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,

  AB = BP,
AB = AE,
AP = EP.

Triangles ABE and AEP are also similar, in particular

  BE / AB = AE / EP, or
BE·EP = AB², i.e.
(BP + EP)·EP = AB², and lastly,
(AB + EP)·EP = AB².

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

  x + 1 = x²,

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(18)

from cos(2α) = 2cos²(α) - 1 and then

  sin(18)

from cos²(α) + sin²(α) = 1. Now, it may be hard to believe but this expression simplifies to

  sin(18°) = (5 - 1)/4,

which is immediately verified by squaring the two expressions. Also

  sin(36)

from sin(2α) = 2sin(α)cos(α).

We can easily find cos(72°) from cos(2α) = 2cos²(α) - 1:

 
cos(72°)= 2cos²(36°) - 1
 = 2[(5 + 1) / 4]² - 1
 = (6 + 25) / 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).

And, of course,

  sin(72)

because sin(72°) = cos(18°).

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

Golden Ratio

  1. Golden Ratio in Geometry
  2. Golden Ratio in an Irregular Pentagon
  3. Golden Ratio in a Irregular Pentagon II
  4. Inflection Points of Fourth Degree Polynomials
  5. Wythoff's Nim
  6. Inscribing a regular pentagon in a circle - and proving it
  7. Cosine of 36 degrees
  8. Continued Fractions

Copyright © 1996-2010 Alexander Bogomolny

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