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Rational Cosine

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By Brad Rodgers on Wednesday, December 26, 2001 - 10:30 pm:

Is there a number such that x is rational, and cos(x) is rational? Since I doubt this is true, is there a number such that x is algebraic and cos(x) is algebraic?

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By Dan Goodman on Wednesday, December 26, 2001 - 11:37 pm:

x=0?

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By Dan Goodman on Wednesday, December 26, 2001 - 11:47 pm:

Sorry for the above post. :-)

I think the less trivial answer is probably no. I seem to remember that there is a theorem that says something like if x is rational and nonzero then e^x is transcendental. If this extends further and we can say that if x is algebraic and nonzero then e^x is transcendental then we're done. Because, if x was algebraic and cos(x) algebraic then also sin(x) is algebraic because sin²(x)+cos²(x)-1=0 and so also is i.sin(x) and i.x. So we have that cos(x)+i.sin(x) is algebraic, i.e. e^ix is algebraic. But this would be a contradiction.

So, does anyone know if the theorem I used is true or not?

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By Michael Doré on Thursday, December 27, 2001 - 12:56 am:

Yes, it's a special case of Lindemann's theorem. Lindemann's theorem says that if a₁,...,a_n are distinct algebraic numbers and if b₁,...,b_n are algebraic then:

b₁e^a₁ + ... + b_ne^a_n = 0

implies that b₁ = ... = b_n = 0.

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By Arun Iyer on Thursday, December 27, 2001 - 09:42 am:

I don't get it why not x=0??

love arun

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By Dan Goodman on Thursday, December 27, 2001 - 02:23 pm:

Arun, x=0 is fine. The point is that x=0 is the only such number. What we've proved above is that there are no other x apart from that obvious one.

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By Arun Iyer on Thursday, December 27, 2001 - 06:11 pm:

Oh!i get it...

love arun

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