Is $\sin n/\sin m$ irrational for nonzero integers $n\neq m$?

Question

As shown in Is sin(x) necessarily irrational where x is rational?, the sine of any rational except for zero is transcendental. But what about the ratio of two such numbers, at least for integers? I thought about expanding $\sin n$ in terms of $\sin 1$ and $\cos 1$, but maybe there's already been some prior work on this?

Answer 1

As shown in the answer you linked, the Lindemann–Weierstrass theorem implies that $\cos 1$ is transcendental. It turns out that $\dfrac{\sin n}{\sin 1}$ is a polynomial in $\cos 1$; in particular, $$\frac{\sin n}{\sin 1}=U_{n-1}(\cos 1)$$ where $U$ is the Chebyshev polynomial of the second kind. Not only that, but $U_{n-1}$ has integer coefficients and has degree $n-1$.

Assume without loss of generality that $0<m<n$. Suppose $\dfrac{\sin m}{\sin n}=\dfrac ab$, with $a,b$ integers. Then dividing numerator and denominator by $\cos 1$, we see that $\dfrac{U_{m-1}(\cos 1)}{U_{n-1}(\cos 1)}=\dfrac ab$, or $$aU_{n-1}(\cos 1)-bU_{m-1}(\cos 1)=0.$$ Since $\cos 1$ is transcendental, it is not the root of any nonzero polynomial with integer coefficients. Thus the polynomial must equal zero everywhere: $$aU_{n-1}(x)-bU_{m-1}(x)=0.$$ But this is impossible as this polynomial has degree $n-1$ and cannot be the zero polynomial.

Answer 2

WLOG we can assume $m,n > 0$. Suppose by absurd that $q = \frac{\sin(n)}{\sin(m)}$ is rational. This would imply $q = \frac{e^{in} - e^{-in}}{e^{im} - e^{-im}}$ and hence $q(e^{im} - e^{-im}) = e^{in} - e^{-in} \Leftrightarrow q(e^{i(2m+n)} - e^{in}) = e^{i(2n+m)} - e^{im}$. So $e^i$ would be a root of $q(x^{2m+n} - x^{n}) - x^{2n+m} - x^{m}$, which can't happen because $e^i$ is trancendental by Lindemann-Weierstrass.

Answer 3

I put it as a comment, but it may be useful to add it here for reference: for every $N \ge 1$ we have that $\sin^2Nx$ is a polynomial of degree $2N$ in $\sin x$ with rational coefficients

(the squaring is needed due to the odd-even split for $N$ since if $N$ odd we have $\sin Nx$ is a polynomial of degree $N$ in $\sin x$ with rational coefficients, but for $N$ even we have that $\sin Nx =\cos x$ times a polynomial of degree $N-1$ in $\sin x$ with rational coefficients which easily follows by induction from the rule for $\sin (x+y)$, so squaring and using $\cos^2 x=1-\sin^2 x$ sidesteps that)

Then applying this to $\sin ^2 mn$ we get that is a polynomial of degree $2m$ in $\sin n$ and one of degree $2n$ in $\sin m$, both with rational coefficients, so if $\sin m =q \sin n, q \in \mathbb Q$, substituting we get that $\sin n$ is a root of a nontrivial polynomial with rational coefficients which contradicts the standard transcendence results for $\sin n, n \ge 1$.

The same argument leads to a contradiction if $q$ is algebraic since roots of polynomials with algebraic coefficients are algebraic too, so the ratio must indeed be transcendental.