Let $f:\mathbb{R}\rightarrow \mathbb{R}^2$ be defined by $f(x)=(\cos(x),\sin(x))$. Does $f(\mathbb{Z})$ cover the unit circle? The word cover is rather vague. To be more precise, is it the case that, given any $r,\epsilon\in\mathbb{R} $ with $f(r)$ not equal to $f(r+\epsilon)$, that there exists some $N\in\mathbb{Z}$ such that 
$$\text{min}(\sin(r),\sin(r+\epsilon))\le \sin(N)\le \max(\sin(r),\sin(r+\epsilon))$$
and
$$\min(\cos(r),\cos(r+\epsilon))\le \cos(N)\le \max(\cos(r),\cos(r+\epsilon))$$
 A: Ok, let us prove that $\{e^{in}\}_{n\geq 1}$ is dense in the unit circle. 
Step1. We prove first that for any $\varepsilon>0$ there are infinite natural numbers such that $\left|e^{in}-1\right|\leq\varepsilon$. We have that $2\pi$ is an irrational number, hence we may extract from its continued fraction a sequence of rational approximations $\left\{\frac{p_m}{q_m}\right\}_{m\geq 1}$ with the property that
$$\left|2\pi-\frac{p_m}{q_m}\right|\leq \frac{1}{q_m^2} $$
implying that $\left|2\pi q_m-p_m\right|\leq \frac{1}{q_m}$. This sequence gives that $1=e^{2\pi i q_m}$ and $e^{ip_m}$ are arbitrarily close, since the sequence $\{q_m\}_{m\geq 1}$ is unbounded and for any $\eta\in\left[0,\frac{1}{4}\right]$ we have $\left|1-e^{2\pi i \eta}\right|=2\sin(\pi\eta)\leq 2\pi\eta$.
Step2. The sequence $\{e^{in}\}_{n\geq 1}$ is dense in the unit circle. If we assume the opposite we have that there is some open interval $J\subset[0,2\pi)$ such that no element of $\{e^{in}\}_{n\geq 1}$ belongs to $J$. How large can such interval be? There are not many chances, indeed. We may fix some $\varepsilon>0$ and find, through the previous step, some $n_0$ such that the length of the arc joining $1$ with $n_0$ is $\leq\varepsilon$. If we consider $e^{in_0},e^{i2n_0},e^{i3n_0},\ldots$ until completing a full turn around the circle, such points break the circle into intervals having length $\leq \varepsilon$. In particular $|J|$ cannot exceed $\varepsilon$, but since $\varepsilon$ can be chosen as small as we like, there cannot be open intervals in $[0,2\pi)$ without any element of $\{e^{in}\}_{n\geq 1}$ in them.
Extra. The sequence $\{e^{in}\}_{n\geq 1}$ is dense in itself. Since $e^{in}\cdot e^{im}=e^{i(n+m)}$ this simply follows from Step1: in order to approximate $e^{in_0}$, it is enough to consider some $e^{iM}$ that is close to $1$. Then $e^{i(n_0+M)}$ is an arbitrarily good approximation of $e^{in_0}$.
