Cos[Im[zetazero(n)]Log(prime)] spans a countable dense set in [-1,1]? It is known that cos(N) spans a countable dense set in [-1,1].
(N: any natural number)
As far as I know generally, for any continuous function f defined in [a,b],
f is Riemann integrable where its domain is a countable dense set in [a,b].
My question: will cos[t_n*Log(p)] Spans a countable dense set in [-1,1]?
*(Variable: n; 1 to infinity)*
t_n=Im[Zetazero(n)]: the imaginary part of the n'th nontrivial zero of the Riemann zeta function.
p: any prime number
 A: For any fixed real $\alpha$, the fractional parts of the numbers $\alpha \gamma$, where $\beta+i\gamma$ runs over all zeros of $\zeta(s)$ in the critical strip  with $0<\gamma < T$, become uniformly distributed in $\mathbf{R}/\mathbf{Z}$ as $T\to \infty$.  This is a theorem of Fujii; see his paper "On the zeros of Dirichlet L-functions, III", Transactions of the AMS, vol. 219.  This affirmatively answers your question.
A: If the cosine values only related to the zeros on the critical line span a countable dense set, every cosine value determined by all the nontrivial zeros can span a countable dense set, too. n can be varied from 1 to infinity, and this I think will make a dense set. Do you agree?
A: Actually a stronger statement holds: $\cos(t_n\cdot\log(p))$ is dense in $[-1,1]$ for $n$ fixed. The idea behind proving this stronger version is to show that for any $x\in[-1,1]$ and $\epsilon>0$ fixed, there is a prime $p$ such that for some $k$ one has: 
$(1)\;\;\;\;\;\;\;\;\;\;\;\;\;|t_n\cdot\log(p)-\arccos(x)-2\pi k|<\epsilon$
The Prime Number Theorem implies $\log(p_{m+1})-\log(p_m)< \frac{1}{2}t_n^{-1}\cdot\epsilon$ if $m$ is large enough (where $p_m$ is the $m^{th}$ prime). Hence for some large enough $k$ one can satisfy $(1)$ with $p$ prime. Since the stronger statement holds, your original statment holds as well.
