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Let $P$ be the set of all prime numbers. Is $\sin(P)$ dense is $[-1,1]?$ How could we approach such a problem?

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    @Gerry: Thanks for correcting that. I can't believe I wrote that and reread it without noticing. Of course, the image under sine of any infinitely long arithmetic progression of integers is dense in$[-1,1]$by the same arguments that work for $\sin(\mathbb{Z})$.2012-02-16

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According to the Wikipedia article about the discrepancy of a sequence:

The sequence of all multiples of an irrational $\alpha$ by successive prime numbers, $2 \alpha$, $3 \alpha$, $5 \alpha$, $7 \alpha$, $11 \alpha$, ... is equidistributed modulo 1. This is a famous theorem of analytic number theory, proved by I. M. Vinogradov in 1935.

With $\alpha = \frac{1}{2 \pi}$, this implies that $P$ is equidistributed modulo $2 \pi$. Using this, and the continuity of the sine function, I think it is straightforward to show that $\sin(P)$ is dense in $[-1,1]$ (although not equidistributed).

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    @draks The phase of $P^{it}$ varies extremely slowly for any fixed $t$, so there's no question that it is dense on the unit circle. One only needs $\frac{p_{n+1}}{p_n} \to 1$, which is slightly stronger than Bertrand's postulate and weaker than the Prime Number Theorem.2013-01-06