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### Trigonometry Without Calculators

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Date: 01/06/97 at 20:15:08
From: Sriram Gollapalli
Subject: Cosine 40

Hi!

I was wondering how to find the cosine of 40 degrees in rational form
and without a calculator.

Thanks, Sriram
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```
Date: 01/07/97 at 07:15:23
From: Doctor Pete
Subject: Re: Cosine 40

Hi,

I am not quite sure what you mean by "in rational form," because
Cos[40] is not a rational number.  If you wish to find an equivalent
expression for it, for example, the way you can by saying that
Cos[60] = 1/2, or Sin[45] = 1/Sqrt[2], then this is also not quite as
simple as it may seem.

In any case, consider the double-angle formula:

Cos[2x] = Cos[x]^2 - Sin[x]^2
= Cos[x]^2 - (1-Cos[x]^2)
= 2 Cos[x]^2 - 1.

Then Cos[40] = 2 Cos[20]^2 - 1.  To find Cos[20], we note:

Cos[3x] = Cos[x+2x]
= Cos[x]Cos[2x] - Sin[x]Sin[2x]
= Cos[x](Cos[x]^2-Sin[x]^2) - Sin[x](2 Sin[x]Cos[x])
= Cos[x]^3 - 3 Cos[x]Sin[x]^2
= Cos[x]^3 - 3 Cos[x](1-Cos[x]^2)
= Cos[x]^3 - 3 Cos[x] + 3 Cos[x]^3
= 4 Cos[x]^3 - 3 Cos[x].

Therefore,

Cos[60] = 4 Cos[20]^3 - 3 Cos[20]
1/2 = 4 Cos[20]^3 - 3 Cos[20]

Hence Cos[20] is the real root of the irreducible cubic

x^3 - 3/4 x - 1/8 = 0

This has the real root:

r = (a^(1/3) + b^(1/3))/2

where a = (1+Sqrt[-3])/2, b = (1-Sqrt[-3])/2 are complex conjugates.

Hence r is the value of Cos[20], and:

Cos[20]^2 = (a^(2/3) + b^(2/3))/4 + 1/2
so
Cos[40] = 2*Cos[20]^2 - 1
= (a^(2/3) + b^(2/3))/2,

where a and b are given as above.

Note that this alternative expression for Cos[40] is indeed a real
number, because a and b are complex conjugates, and if the appropriate
cube root is taken, these will also be conjugates, and hence their
imaginary parts will cancel.  There is no way of avoiding the usage of
complex numbers in solving for the roots of the associated cubic for
Cos[20].  Note that this shows that the regular polygon (a 9-gon) with
interior angles of size Cos[40] is not constructible by Euclidean
methods of straightedge and compass, because Cos [40] (the size of the
angles you would need to draw to construct the polygon) is the
solution of an irreducible cubic.

If you are wondering how I obtained the real root r (you may
substitute it and easily verify that it is indeed a root), it was
found using Mathematica, a powerful symbolic algebra computer program.
However, it is just as possible to find the solution to the cubic
using pencil-and-paper methods.

One last note:  It can be shown that

Cos[72] = 1/(1+Sqrt[5]),

hence the regular pentagon is constructible by straightedge and
compass.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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