Find the number of all natural solutions to $$0<\sin n < 10^{-12}, \ \ n\in \mathbb{N}$$
My try: Consider a unit circle on $xy$-plane. Then all values of $\sin n$ can be illustrated by rotating the point $(\cos1,\sin1)$ by an angle of value $1$ radian $(\sin 2, \sin 3, ...)$. Converting this into degree, one gets $\alpha = 180^\circ/\pi$. Since $\alpha$ is irrational, this rotation process cannot be periodic (never coincides with the initial step). Therefore, points of the form $(\cos n,\sin n)$ are dense on the circle, which, particularly, provides that there are infinitely many values of $n \in \mathbb{N}$ satisfying $0<\sin n < 10^{-12}.$
Is there anything I am missing for complete proof?
Update: Below is my try to complete, taking into note the comments and the answer:
Let $$T_\alpha^{(n)}(x) = x + n\alpha \mod 1$$, where $x$ - initial point, $n$ - number of rotations. In our case $\alpha$ is irrational.
- We should show that all $T_\alpha^{(n)}(x)$ are distinct. Assume that $\exists k,i,j:$ $$T_\alpha^{(i)}(x) = T_\alpha^{(j)}(x) + k, \ \ i \neq j$$ $$x+i\alpha = x+ j\alpha + k \implies (i-j)\alpha = k$$ Since $\alpha$ is irrational, we must have $i=j$. Contradiction.
Now, divide $[0,1]$ into $n$ parts, length of $1/n$ each. We know that $x, T_\alpha^{(1)}(x),...,T_\alpha^{(n)}(x)$ are distinct. Thus, by Drichlet's principle, there exists an interval that contains at least two of the terms of the sequence above. Say $i^{th}$ and $j^{th}$, $(i \neq j)$ $$|T_\alpha^{(i)}(x)-T_\alpha^{(j)}(x)|<\frac{1}{n}$$
- We should prove that $\forall x_0 \in [0,1], \forall \varepsilon > 0, \exists m \in \mathbb{Z}:$ $$T_\alpha^{(m)}(x) \in \bigcup_\varepsilon(x_0)$$ We know that $$\frac{1}{n} > |T_\alpha^{(i)}(x) - T_\alpha^{(j)}(x)| = |T_\alpha^{(i-j)}(x)-x|$$ Thus, taking $m:=i-j$, we have $$T_\alpha^{(m)}(x) \in \bigcup_\varepsilon(x_0)$$
Therefore, $T_\alpha^{(n)}(x)$ is dense at $[0,1]$.
Hint Use the irationality of $\pi$ to show that the set $$\{ n+2m \pi : n \in \mathbb {N}, m \in \mathbb{Z} \}$$ is dense in $\mathbb R$.