Transcendental Values for sin(Q*pi)

1. Jul 30, 2010

marschmellow

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. Jul 30, 2010

Hurkyl

Staff Emeritus
Multiple angle formula. Or complex exponential, I suppose.

3. Jul 30, 2010

marschmellow

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.

4. Jul 31, 2010

hamster143

if $\sin{Q \pi} = x$, where Q is rational, by the virtue of the multiple angle formula, there is a polynomial P(x,y) with integer coefficients such that $P(\sin{Q \pi}, \cos{Q \pi}) = P(x, \sqrt{1-x^2}) = 0$.

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

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

Therefore x is not transcendental.

5. Jul 31, 2010

marschmellow

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.

6. Aug 1, 2010

hamster143

Do you remember the definition of "transcendental"? Can you write down formulas for sin(2x) and sin(3x) and take it from there?

7. Aug 1, 2010

Let $A = \sin(\pi a/b)$, where $a,b$ are integers. Then $p(A) = \sin(\pi a) = 0$ for some polynomial $p$ with rational coefficients, by using an appropriate multiple angle formula. Since $A$ 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 $A = \sin(\pi a/b) = (x + \bar x)/2i$, where $x = e^{i\pi a/b}$. Now $x^{2b} - 1 = 0$, proving that $x$ is algebraic, and thus so is $A$.

Last edited: Aug 1, 2010
8. Aug 3, 2010

marschmellow

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.