Suppose $\cos(t) $ and $\sin(t) $ are rational but not in $\{-1,0,1\}$. Since the Gaussian integers are a UFD, we can write $z = \exp(it) = \cos(t) + i \sin(t) = \prod_j p_j^{d_j}$ where $p_j$ are Gaussian primes and $d_j$ are integers, with finitely many (and at least one) $d_j$ nonzero.
Similarly, if $s = mt/n$ for coprime integers $m,n$ and $\cos(s)$ and $\sin(s)$ are rational, write
$w = \exp(is) = u\prod_j p_j^{e_j}$ where $u \in \{\pm 1, \pm i\}$. Now $w^n = z^m$, so $d_j m = e_j n$ for all $j$. In particular, $m/n = e_j/d_j \in {\mathbb Z}/d_j$. We conclude that for all nonzero $c$, there is at most a discrete set of rational points on the spiral.

EDIT: In fact, we don't need the $y$ component: for all nonzero $c$, $\{(ct, \cos(t)): t \in {\mathbb R}\}$ has only a discrete set of rational points.
Note that $\cos(nt) = T_n(\cos(t))$ where $T_n$ is the $n$'th Chebyshev polynomial of the first kind. These satisfy the recursion
$T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)$ with $T_0(x) = 1$ and $T_1(x) = x$.

It's easy to prove by induction that for any odd prime $p$, if the $p$-adic
norm $|x|_p = q > 1$, then $|T_n(x)|_p = q^n$, while if $q \le 1$, $|T_n(x)|_p \le 1$. On the other hand, for $p=2$, if $|x|_2 = q > 2$ then $|T_n(x)|_2 = 2 (q/2)^n$ while if $|x|_2 \le 2$ then $|T_n(x)|_2 \le 2$.

Now suppose $\cos(t) = x$ and $\cos(s) = w$ are nonzero and rational and $s/t = m/n$ is rational, where $m$ and $n$ are positive integers.
We have $T_n(w) = \cos(ns) = \cos(mt) = T_m(x)$. Fix $t$ and take $0 < |m/n-1| < \epsilon$ where $\epsilon$ is small. If $\epsilon$ is sufficiently small, there is some prime $p$ such that $|w|_p \ge \max(p,|x|_p^2)$ (if $p \ne 2$) or
$\max(4, |x|_2^2)$ (if $p = 2$).
But then (in the case $p > 2$)
$|T_n(w)|_p = |w|_p^n \ge \max(p^n,|x|_p^{2n}) > |T_m(x)|_p$

or (in the case $p=2$)
$|T_n(w)|_2 = 2 (|w|_2/2)^n \ge \max(2^{n+1}, 2 (|x|_2^2/2)^n) > |T_m(x)|_2$.
The contradiction shows that $s$ can't be taken too close to $t$.

all$c$, for cardinality reasons; indeed for all but countably many $c$ there's only the obvious point $(t,x,y,z) = (0,1,0,0)$. [If $t \neq 0$ then $c = (\cos^{-1}x) / z$ with countably many $(x,z)$ pairs each of which has countably many choices of $\cos^{-1} x$.] If we get to choose $c$ then there can be infinitely many rational points, but still not a dense set, only a discrete subgroup. $\endgroup$ – Noam D. Elkies Sep 8 '13 at 1:08Mathematica. $\endgroup$ – Joseph O'Rourke Sep 8 '13 at 20:44