Rational values of $\sin(\log(x))$

Question

Apart from the trivial solution $\sin(\log(1))=0$, is

$$\sin(\log(x))$$

ever rational if $x$ is rational?

Answer 1

We will show more: that if $x \ne 0, 1$ is algebraic (possibly non-real), then $\sin(\log(x))$ is transcendental (regardless of which value of the logarithm we take). To do this, we apply the Gelfond-Schneider Theorem, which states that if $\alpha, \beta$ are algebraic numbers with $\alpha \ne 0,1$ and $\beta \notin \mathbb{Q}$, then any value of $\alpha^\beta$ is transcendental. (See Ivan Niven's book, Irrational Numbers, for proof of this and related results.)

Suppose $x \in \mathbb{C}$ is algebraic and not 0 or 1. Let y = $\log x$ (where we pick once and for all a particular value of $\log x$) and $z = \sin y$. Then $$z = \frac{e^{iy} - e^{-iy}}{2i},$$ so $e^{iy} - e^{-iy} - 2zi = 0$. Multiplying by $e^{iy}$, we have $(e^{iy})^2 - 1 - 2zie^{iy} = 0$. Substituting $y = \log x$, we obtain$$(x^i)^2 - 2zix^i - 1 = 0.$$

By the Gelfond-Schneider Theorem, $x^i$ is not algebraic, and hence is not the solution of a non-zero polynomial with algebraic coefficients. Hence, we must have $z$ not algebraic.