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Transcendental Values for sin(Q*pi)

  1. Jul 30, 2010 #1
    Simple question that for some reason I can't reason myself through. I'm probably gonna be mad at myself when someone provides me with the insight needed to solve this and I didn't find it on my own, but anyway, here goes:

    If sin(Q*pi)=A, where Q is rational, can any values of A be transcendental?
     
  2. jcsd
  3. Jul 30, 2010 #2

    Hurkyl

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    Multiple angle formula. Or complex exponential, I suppose.
     
  4. Jul 30, 2010 #3
    Thanks for the help, but that only covers integer values of Q, I think. If Q is restricted to integers then sin(Q*pi)=0.
     
  5. Jul 31, 2010 #4
    if [itex]\sin{Q \pi} = x[/itex], where Q is rational, by the virtue of the multiple angle formula, there is a polynomial P(x,y) with integer coefficients such that [itex]P(\sin{Q \pi}, \cos{Q \pi}) = P(x, \sqrt{1-x^2}) = 0[/itex].

    Furthermore, [itex]P(x, \sqrt{1-x^2}) = R(x) + S(x) \sqrt{1-x^2}[/itex] for some integer polynomials R and S.

    Therefore there is an integer polynomial [itex]R(x)^2 - S(x)^2(1-x^2) = 0[/itex] that has x as its root.

    Therefore x is not transcendental.
     
  6. Jul 31, 2010 #5
    I can't say I completely followed that, but I believe you, and I'm honestly more interested in the answer than an explanation, however weird that is for math questions.
     
  7. Aug 1, 2010 #6
    Do you remember the definition of "transcendental"? Can you write down formulas for sin(2x) and sin(3x) and take it from there?
     
  8. Aug 1, 2010 #7
    Let [itex]A = \sin(\pi a/b)[/itex], where [itex]a,b[/itex] are integers. Then [itex]p(A) = \sin(\pi a) = 0[/itex] for some polynomial [itex]p[/itex] with rational coefficients, by using an appropriate multiple angle formula. Since [itex]A[/itex] is the root of a polynomial with rational coefficients, it is not transcendental.


    edit: I've spent some time thinking about it (read: sleep), and have come up with a nicer solution. Observe that [itex]A = \sin(\pi a/b) = (x + \bar x)/2i[/itex], where [itex]x = e^{i\pi a/b}[/itex]. Now [itex]x^{2b} - 1 = 0[/itex], proving that [itex]x[/itex] is algebraic, and thus so is [itex]A[/itex].
     
    Last edited: Aug 1, 2010
  9. Aug 3, 2010 #8
    Wow, thanks a lot. That explanation makes sense, but I couldn't have thought of it on my own, so I'm glad I have no regrets for asking for help.
     
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