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.
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,
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AB = BP,
AB = AE,
AP = EP.
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Triangles ABE and AEP are also similar, in particular
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BE / AB = AE / EP, or
BE·EP = AB², i.e.
(BP + EP)·EP = AB², and lastly,
(AB + EP)·EP = AB².
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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
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cos(∠AEP) = (AE/2)/EP = (AE/EP)/2 = φ/2.
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But ∠AEP = 36° and we get the desired result.
Using cos(36°) = (1 + √5)/4 we can find
from cos(2α) = 2cos²(α) - 1 and then
from cos²(α) + sin²(α) = 1. Now, it may be hard to believe but this expression simplifies to
which is immediately verified by squaring the two expressions. Also
from sin(2α) = 2sin(α)cos(α).
We can easily find cos(72°) from cos(2α) = 2cos²(α) - 1:
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| cos(72°) | = 2cos²(36°) - 1 |
| | = 2[(√5 + 1) / 4]² - 1 |
| | = (6 + 2√5) / 8 - 1 |
| | = (3 + √5) / 4 - 1 |
| | = (√5 - 1) / 4. |
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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,
because sin(72°) = cos(18°).
Trigonometry
Fibonacci Numbers
- Ceva's Theorem: A Matter of Appreciation
- When the Counting Gets Tough, the Tough Count on Mathematics
- I. Sharygin's Problem of Criminal Ministers
- Single Pile Games
- Take-Away Games>
- Number 8 Is Interesting
- Curry's Paradox
- A Problem in Checker-Jumping
- Fibonacci's Quickies
Golden Ratio
- Golden Ratio in Geometry
- Golden Ratio in an Irregular Pentagon
- Golden Ratio in a Irregular Pentagon II
- Inflection Points of Fourth Degree Polynomials
- Wythoff's Nim
- Inscribing a regular pentagon in a circle - and proving it
- Cosine of 36 degrees
- Continued Fractions
Copyright © 1996-2010 Alexander Bogomolny
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