Transcendental Values for sin(Q*pi)

[marschmellow]

Simple question that for some reason I can't reason myself through. I'm probably going to 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?

 

[Hurkyl]

Multiple angle formula. Or complex exponential, I suppose.

 

[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.

 

[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.

 

[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.

 

[hamster143]

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

 

[adriank]

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 $A$.

 

[marschmellow]

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 $A$.

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.