**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 10**º, **sin 50**º
**and
*minus*
sin 70**º**.

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:

(2R**² **- M**²**)/R**²
**=
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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