Sin(10) is an irrational number
David McNaughton

As explained in DLMcN.com/sin10-surd.html, the roots of
8t³ - 6t + 1 = 0 ... <1>
are sin 10º, sin 50º and minus sin 70º.

These three quantities also have a "circular", interlocking relationship determined by the standard formula
cos 2A = 1- 2sin²A
... giving us
cos 20 = sin 70 = 1 - 2sin²10 ... <2>
cos 140 = -sin 50 = 1 - 2sin²70 ... <3>
cos 100 = -sin 10 = 1 - 2sin²50 ... <4>
- demonstrating that if sin 10 is rational, then sin 70 and sin 50 are too.

Let us therefore assume that they are all rational. If the result is unsustainable nonsense, then that will prove that they are not rational numbers.

Suppose that positive integers do exist (I, J, K, L, M and N) for which
sin 10 = M/N ... <5>
sin 50 = I/J ... <6>
sin 70 = K/L ... <7>
where these fractions have been reduced to their simplest forms, such that there is no common factor between numerator and denominator (discounting 1, obviously).

Equation <1> is equivalent to
(t - sin 10)(t - sin 50)(t + sin 70) = 0
which, if our assumptions are true, becomes
(t - M/N)(t - I/J)(t + K/L) = 0 ... <8>

Multiplying out and comparing the "constants" in <8> and <1> gives (as the Product of the Roots):
IKM/JLN = 1/8 ... <9>

We may also compare the coefficients of t², giving (as the Sum of the Roots):
M/N + I/J - K/L = 0... <10>
- although I have not actually made use of this equation in this article, nor indeed the other relationship obtained by comparing the coefficients of t.

Rewrite <2> as
(N² - 2M²) / N² = K/L ... <11>
and <3> as
(2K² - L²) / L² = I/J ... <12>
and <4> as
(2I² - J²) / J² = M/N ... <13>

In <11>, if N is an odd number, L will also have to be odd (and indeed K too) - and then from <12>, so will J and I. With those requirements, equation <9> could never be fulfilled - because it would be impossible to obtain a value of 8 in the denominator.

So, remembering that <5>, <6> and <7> contain "reduced" fractions, they must all take the form "odd/even".

Under those circumstances, if N is a multiple of 4, then an examination of <11> shows that L has to be a multiple of 8. If that is so, consideration of <12> establishes that J would then be a multiple of 32. And feeding that information into <13> requires N to be a multiple of 512. We can obviously continue like that indefinitely, "circling" through the sequence <11>, <12>, <13> and back to <11>, eventually deducing that N, L and J must be multiples of enormous numbers - which continue to grow with absolutely no upper limit. So N cannot be a multiple of 4, (nor can L or J be - because we could just as easily have started our discussion with these). By halving N, L or J we therefore obtain an odd number; (this leads to the note appended at the end of this article).
 

Now let us see what happens if we postulate that N is a multiple of 3. M cannot then contain 3 as a factor (because M/N is a reduced fraction), and an evaluation of <11> [similar to that in the last paragraph] demonstrates that L will then be a multiple of 9. Continuing to <12>, J would need to be a multiple of 81 - and (carrying on "round the circle" to <13>) we would be forced to conclude that N was a multiple of 6561 etc. etc. ... once again with no "ceiling" on the possibilities. So N cannot be a multiple of 3, nor can L or J be.

That reasoning involving multiples of 3 - may now be repeated for every prime number, no matter how large. Thus, no finite number can ever fulfil the role of J, L or N, so the sine of 10 degrees must be irrational.
--------------------------------------------------------

Knowing that J, L and N were not multiples of 4, we could of course have written
J = 2P ... L = 2Q ... and N = 2R
- enabling us to work with odd-no./odd-no. combinations like K/Q rather than K/L, etc.

For example, equation <11> would then have yielded:
(2R² - M²)/R² = K/Q, with all values being odd integers. This might have made it easier to deduce that those integers cannot contain any prime factors other than 2; in other words, such numbers simply cannot be found to represent and express sin 10.


A much easier proof is available - although we need to appeal to the Rational Root Theorem - [originally placed in http://en.wikipedia.org/wiki/Rational_root_theorem ... which is a special case of Gauss's Lemma on the factorization of polynomials].

That alternative approach becomes even more straightforward if we rewrite <1> using
y = 2t ...

... enabling us to work with
y³ - 3y + 1 = 0


David McNaughton
E-mail: DLMcN@yahoo.com

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