## Sin(q*pi)=r; q,r rational

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aguacate
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### Sin(q*pi)=r; q,r rational

I got started on this problem while traveling down on a tangent to my algebra homework. Let Q be the rationals and R be the reals. Define f:Q->R by f(q)=Cos(q*pi). How many elements are in f(Q) \cap Q?

(by \cap I mean intersect).

edit: Yes I did mean "f(Q) \cap Q" and not "f(Q) \cap R"
Last edited by aguacate on Fri Feb 22, 2008 5:19 pm UTC, edited 1 time in total.

Owehn
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### Re: Sin(q*pi)=r; q,r rational

Did you mean f(Q)\cap Q? My guess would be just -1, 0, and 1.

Edit: Bother, sorry I forgot ±1/2.
Last edited by Owehn on Fri Feb 22, 2008 1:00 am UTC, edited 1 time in total.
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CodeLabMaster
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### Re: Sin(q*pi)=r; q,r rational

If q = (1/6), then f(q) = 1/2, so there are definitely more than just {-1,0,1}. Unless I'm mistaken about the nature of your question, shouldn't there be an infinite amount of elements?

Cosmologicon
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### Re: Sin(q*pi)=r; q,r rational

ISTR that cos(pi x) is algebraic if and only if x is rational, in which case the answer would be all the rationals in [-1,1] are in f(Q).

adlaiff6
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### Re: Sin(q*pi)=r; q,r rational

To find the rational points on a curve, you need to first find one such point, (a,b). Now, parameterise the line y-b = m(x-a) such that m is rational. Now, if x is rational, y must also be rational, so every point with a rational x coordinate on this line will also have a rational y coordinate. If you now find the points where that parameterised line intersects your curve (at a rational x), that will correspond to a rational point on your curve.

Try this out with your problem and see if you get anywhere.
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NathanielJ
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### Re: Sin(q*pi)=r; q,r rational

The answer is {-1, -1/2, 0, 1/2, 1}, but to get that I had to use some complex analysis. I'll post a solution later tonight if no one else has gotten around to it.
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Ended
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### Re: Sin(q*pi)=r; q,r rational

Cosmologicon wrote:ISTR that cos(pi x) is algebraic if and only if x is rational

Hmm, the 'if' direction is fairly easy to prove:

Spoiler:
In general, we know cos(nx) = Tn(cos(x)) for a polynomial Tn.
So cos(x) = polynomial in cos(x/n) in general. (1)

Let x = p/q. Have cos(p pi) = integer.
And cos(p.pi) = polynomial in cos(p/q . pi) from (1)

Therefore cos(p/q . pi) is algebraic.

The 'only if' direction (which we need) seems much harder though. [edit: NathanielJ's post suggests it's not true]
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NathanielJ
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### Re: Sin(q*pi)=r; q,r rational

Kk here's the sketch of a proof of my claim:

First, write r = m/n and 2cos(r*pi) = e^(i*pi*m/n) + e^(-i*pi*m/n). Using the binomial expansion theorem, you can see without too much trouble that 2cos(r*pi) is the root of some monic polynomial of degree 2n with integer coefficients (though actually coming up with the coefficients would be ugly as sin). Thus, 2cos(r*pi) is an algebraic integer (see http://en.wikipedia.org/wiki/Algebraic_integer and note that the term "algebraic integer" is a bit of a misnomer, as algebraic integers aren't necessarily integers). However, the only algebraic integers that are rational are actually normal integers, so 2cos(r*pi) must be an integer, from which it immediately follows that cos (r*pi) is one of -1, -1/2, 0, 1/2, or 1.
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