Is $\sin(e)$ rational or irrational?

Solution 1:

** comment **
Mark says to use Schanuel's conjecture. Here it is.

We know $e$ is real and nonzero, so the two numbers $ie$ and $1$ are linearly independent over the rationals. From Schanuel's conjecture, we conclude that the transcendence degree of $$ \mathbb Q(ie,1,e^{ie},e^1) $$ is at least $2$. But $1$ is algebraic, and $ie, e$ are algebraically related, so we conclude that $e^{ie}$ is transcendental. Now $$ \sin e = \frac{e^{ie}-e^{-ie}}{2i} = \frac{1}{2i}\left(e^{ie} - \frac{1}{e^{ie}}\right) $$ so if $\sin e$ were algebraic, then solving a quadratic equation we would conclude that $e^{ie}$ would be algebraic. Therefore, $\sin e$ is transcentental.

Of course, Schanuel's conjecture is only a conjecture...