For $x \notin \pi\mathbb Q$, that is, a real $x$ that is not a rational multiple of $\pi$, consider the set $$\{(\cos nx,\sin nx):n = 0,1,2,…\}.$$ It is known that this set is dense in the unit circle $B(0,1)$ of $\mathbb R^2$. Could someone please give me a proof or reference for a proof?
From $x\notin \pi{\mathbb Q}$ it follows that $e^{ikx}\ne1$ for all $k\in{\mathbb Z}\setminus\{0\}$, and this implies that the numbers $e^{inx}\in S^1$ $(n\geq0)$ are all different. Assume that a point $\zeta\in S^1$ and an $\epsilon>0$ is given. Since $S^1$ has finite length we then can find two numbers $n_1<n_2$ with $|e^{in_2x}-e^{in_1 x}|< \epsilon$. Put $n’:=n_2-n_1$. Then
$$\bigl|e^{in’x} -1\bigr|=|e^{in_2x}-e^{in_1 x}|<\epsilon\ .$$
Put $n_k:=k\>n’$ $(k\geq0)$. Then the successive points $e^{i n_k x}\in S^1$ $(k\geq0)$ have a distance $<\epsilon$. It follows that there is a $k\geq0$ with $|e^{in_k x}-\zeta|<\epsilon$.
As I noted in the comments many Dynamical Systems book will have a proof. I think the book by Robert L. Devaney, An Introduction to Chaotic Dynamical Systems does, but I don’t have access to it right now to confirm. The proof which I enclose below (and which I remember having seen before, perhaps on other books) is from Introduction to Dynamical Systems by Michael Brin, Garrett Stuck (link to 1-st edition at Amazon, though I know there are newer editions).
There are links to this book online, in particular there is a sample of the first chapter at a Library of Congress web page (so I assume it is legal and more or less permanent). See section 1.2, Circle Rotations, p.3-4 there. To make this answer self contained I enclose the proof from the Brin/Stuck book (with some editing, omitting the subscript $\alpha$ and a certain inequality, and comments).
For
$\alpha\in\mathbb R$, let $R$ be the rotation of
the circle $S^1$ by angle $2\pi\alpha$.
If $\alpha=\frac pq$ is rational, then
$R^q=Id$, so every orbit is periodic. On the
other hand, if $\alpha$
is irrational, then every positive semiorbit is dense in
$S^1$.
Indeed, the pigeon-hole principle implies that, for any
$\varepsilon>0$, there are $m,n$ such that
$m<n$ and $d(R^m,R^n)<\varepsilon$.
Thus $R^{n-m}$ is rotation by an angle
less than $\varepsilon$, so every positive semiorbit is
$\varepsilon$-dense in $S^1$ (i.e., comes within
distance $\varepsilon$ of every point in $S^1$). Since
$\varepsilon$ is arbitrary, every positive semiorbit is dense.
Comment. To apply the pigeon-hole principle, you may represent the circle as the union of finitely many closed arcs, each of length less than $\varepsilon$. Then (for any fixed $x$) we place infinitely many $R^n(x)$, $n\ge0$, into finitely many arcs, hence at least one arc must contain both $R^n(x)$ and $R^m(x)$ for some $n\not=m$.
More precisely the book says “for any $\varepsilon>0$, there are $m,n<\frac1\varepsilon$ such that $m<n$” etc, but I do not see why this inequality would be relevant (I first thought it was a typo and should be $m,n>\frac1\varepsilon$ which is equivalent to $\frac1m,\frac1n<\varepsilon$ and implies $|\frac1m-\frac1n|<\varepsilon$), though it may have something to do with an attempt to find $n,m$ that work, with $n>m$ and smallest possible $n-m$. Or it may have something to do with the other representation of rotation (as indicated on the same page), as the sum $\mod1$ on $[0,1]$ with $0\sim1$.
It should be noted in the above proof that for irrational $\alpha$ (i.e. when the angle of rotation $2\pi\alpha$ is an irrational multiple of $\pi$), then $R^n$ is never the identity, for $n\ge1$. Indeed, if $R^n=Id$, then $n2\pi\alpha=2\pi k$ for some integer $k$, hence $\alpha=\frac kn$ would be rational.
Edit. The proof is available in Devaney’s book too, e.g. in the second edition it is Example 3.12 and Theorem 3.13 (referred to as Jacobi’s Theorem there) on p.22-23. He considers the forward orbit of a point and uses that any infinite set (in particular the forward orbit) must have a limit point (and hence, in the notation of the present answer, $d(R^m,R^n)$ could be made arbitrarily small, non-zero, for suitable $m,n$).
You can proove that every rational number other than 1/2 comes from an irrational angle.
a = cos(pi d/z) + i sin(pi d/z) solves some equation of the form a^z + 1 = 0
Prove that a can never be a rational, by supposing that where
$$a^n = \sum^{z-1}_{e=0} x_i a^e$$
that if a were a fraction p/q, multiply through by p^n, then the LHS is not a multiple of q, and the RHS is, unless q=1.
Therefore the equation a can never be rational unless it is half-integer, which means that the rational numbers arise from irrational angles.
Since the rational numbers are dense on the circle, the irrational angles are too.
Suppose $[a,b] \subset [0,1]$. Since $T\theta$ is ergodic,
$$\lim_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} \chi_{[a,b)}(T_\theta ^n (t))=b-a $$