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I'm considering the set $$\left\{\sin(k)\mid k\in\Bbb Z\backslash \left\{0\right\}\right\}.$$ All of its members are transcendental numbers, but together they don't form the complete set of all transcendental numbers between $-1$ and $1$.

Does this set of numbers belong to, and form on $\left[-1,1\right]$ completely, a differently named class of numbers? What would that name be?

Edit due to comment. Or, if not, would these numbers belong to (but not completely form between $-1$ and $1$) a differently named subclass of transcendental numbers?

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It seems unlikely that something so specific would have its own name. If you want shorter notation you could just call it $\sin(\mathbb{Z}\setminus\{0\})$. – Matthew Pressland Mar 7 '13 at 15:27
All of these numbers are elementary numbers in the sense of Timothy Y. Chow. Also, it's probably known within fairly narrow bounds what their transcendence measures are (probably all their transcendence measures are the same), but I don't know much about this. However, none of these properties will characterize the set of numbers you're interested in, and I doubt there is any interest in such a characterization. Perhaps more interesting would be to obtain such a characterization for the values of $e^z$ when $z=x+iy$ with $x,y \in {\mathbb Q}.$ – Dave L. Renfro Mar 7 '13 at 16:36
up vote 1 down vote accepted

there is no name for this class. You will end up with $\aleph_0$ items in the set which is dense in $(-1,1)\cap\mathbb{R}$. In fact, proving that this is dense is going to be hard.

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Actually the fact that they are dense in this interval follows pretty much immediately from the Weyl equidistribution theorem. – Peter Humphries Mar 20 '13 at 19:10

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