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Proof That sin(5) is Irrational


Date: 04/24/2001 at 06:57:20
From: Howard
Subject: sin 5

How do you prove that sin(5) is an irrational number?


Date: 04/24/2001 at 09:52:44
From: Doctor Floor
Subject: Re: sin 5

Hi, Howard,

Thanks for writing.

I suppose you mean that 5 is 5 degrees here. If you mean 5 radians, 
then the situation is a lot more complicated, and I would not know how 
to solve your problem.

We know that 5 degrees is a solution of sin^2(9x) = sin^2(45) = 0.5. 

We try to find a formula for sin(9x) in terms of sin(x). In fact we 
are only interested in the highest power term. By that we could 
rewrite sin^2(9x) = 0.5 into an equation of the form f(sin x) = 0 to 
which we will apply the rational root theorem.

First for sin(3x) we write:

     sin(3x) = sin(x + 2x) 
             = sin(x)cos(2x) + cos(x)sin(2x)
             = sin(x)(1-2sin^2(x)) + cos(x)*2sin(x)cos(x)
             = sin(x) - 2sin^3(x) + 2sin(x)cos^2(x)
             = sin(x) - 2sin^3(x) + 2sin(x)*(1 - sin^2(x))
             = sin(x) - 2sin^3(x) + 2sin(x) - 2sin^3(x)
             = -4sin^3(x) + 3sin(x)

From this we derive:

     sin(9x) = -4sin^3(3x) + 3sin(3x)
             = -4(-4sin^3(x) + 3sin(x))^3 -4sin^3(x) + 3sin(x)

so that

     sin(9x) = -256sin^9(x) + integers times lower powers of sin(x).

Note that the right-hand side does not have a constant term.

This means that sin^2(9x) = 0.5 can be rewritten as:

     65536sin^18(x) + integers times lower powers of sin(x) - 0.5 = 0
   131072 sin^18(x) + integers times lower powers of sin(x) - 1   = 0
     2^17 sin^18(x) + integers times lower powers of sin(x) - 1   = 0

This is a polynomial equation in sin(x) to which we can apply the 
rational root theorem, and sin(5) is one of the roots.

For the rational root theorem, see for instance from the Dr. Math 
archives:

   Rational Root Theorem
   http://mathforum.org/dr.math/problems/drummond8.27.98.html   

We find that the rational roots of this equation must be of the form 
+- (1/2)^n for n <= 17.

It is easy to verify that sin(5), approximately equal to 0.087, is not 
of this form. So sin(5) is a root of the polynomial, but not a 
rational root. That means that sin(5) must be irrational.

I hope this helps. If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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