Numbers of the form sin(

*n*) get arbitrarily close to every real value from -1 to 1.
Consider the
integers mod(2pi). Equivalently, we can
consider the points (cos

*n*, sin*n*) on the unit circle (a.k.a. using the "wrapping function"). Because we have an infinite set of points in a compact set, there must be an accumulation point.
For any epsilon >
0, we can find distinct integers

*m*and*n*such that (cos*m*, sin*m*) and (cos*n*, sin*n*) are less than epsilon apart on the circle, so (cos(*m*-*n*), sin(*m*-*n*)) is within epsilon of (1,0). Then (by say the Archimedean principle), every point on the unit circle is within epsilon of some integer multiple of (*m*-*n*) wrapped around the unit circle . Thus the points of the form (cos*k*, sin*k*) form a dense subset of the unit circle. In particular, numbers of the form sin(*k*) form a dense subset of [-1,1].
A slightly different
argument for density of sin(

*n*) depends on the following lemma.
Lemma: If

*x*is irrational, then the additive group Z +*x*Z is dense in R.
If we accept the
lemma, then the additive group Z + (2pi)Z is dense in R, so the points (cos

*t*, sin*t*) for*t*in Z + (2pi)Z must be dense in the unit circle. But (cos (*n*+m2pi), sin (*n*+m2pi)) = (cos*n*, sin*n*), so (cos*n*, sin*n*) is dense in the circle.
To prove the lemma,
consider the contrapositive and assume that Z +

*x*Z is discrete. Then Z +*x*Z must be cyclic: Z +*x*Z =*d*Z for some real number*d*. Because Z is contained in*d*Z, there must be some integer*n*so that 1 =*nd*, hence*d*= 1/*n*is rational. And then for some*m*, 1+*x*=*m*/*n*, so*x*is rational.
And the converse of
the lemma is also true, because of

*x*=*m*/*n*is rational, then Z + xZ = 1/*n*Z.