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Trigonometry Without Calculators

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


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

Thanks, Sriram

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


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].


     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
     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 

-Doctor Pete,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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