What is the infimum? I'm trying to find  $\inf_{n \in \mathbb N} (\sin(n))^2 $. I think that the answer is $0$ but I couldn't prove it. I appreciate any help. 
 A: Theorem: Let $a>0$ be irrational. Then the sequence of natural numbers is dense in $\mathbb R/(a\mathbb Z)$.
Proof:
The sequence is injective: Assume $n$ and $m$ with $n\ne m$ are equivalent in $\mathbb R/(a\mathbb Z)$. That means $n-m=az$ for a non-zero integer $z$. So $a=(n-m)/z \in\mathbb Q$. Contradiction.
Since $\mathbb R/(a\mathbb Z)$ is compact and the sequence infinite (implied by injectivity) it has a cluster point. In particular we have $n,m$ with $n<m$ which are arbitrarily close together. Let $b\in \mathbb R/(a\mathbb Z)$ and $\varepsilon>0$. Choose $n,m$ with distance less then $\varepsilon$. Then by adding multiplies of $(m-n)$ to an arbitrary element of $\mathbb R/(a\mathbb Z)$ we get into an $\varepsilon$-neighborhood of $b$.
QED.
Answer of the question:
Because $\pi$ is irrational the sequence of natural numbers is dense in $\mathbb R/(\pi\mathbb Z)$ (by our Theorem). So there is an increasing sequence of natural numbers $(a_n)_{n\in\mathbb N}$ converging to the equivalence class of $0$. Since $\sin(x)^2:\mathbb R\to \mathbb R$ is periodic of period $\pi$ it factors through $\mathbb R/(\pi\mathbb Z)$. Continuity of $\sin(x)^2$ gives you
$$\lim_{n\to\infty }\sin(a_n)^2 = 0.$$
Since we have $\sin(x)\ge 0$ for all $x\in\mathbb R$ we have shown
$$\inf_{n\in\mathbb N} \sin(n)^2=0.$$
A: Rather than answering the question entirely, let me give a hopefully helpful suggestion:
Say you want to show that for some $n$, $\sin^2n<\epsilon$ for some positive $\epsilon$. Draw the unit circle! What you want to say is that for some $n$, the point "$n$ radians" lies within $\sqrt{\epsilon}$ of the $x$-axis. Essentially, it would be enough to show that if you plot the integer-radian points on the unit circle, they would basically "fill" it in some sense. Can you do this? (HINT: consider the infimum of the set of distances between integer-radian points, and use the fact that $\pi$ is irrational.)
A: Hint: There is a theorem that says that for any non-negative number $\alpha$, you can find arbitrarily large integers $p,q$ such that $|\alpha - p/q| \leq 1/q^2$. If you combine this with a theorem about how to compare $|\sin x|$ to $|x|$ for small $|x|$, you'll show the answer is indeed $0$ just as you thought.
A: Assuming you already know that $\sin$ is continuous and $2\pi$ periodic.
Let $\epsilon > 0$. Then, there exists some $\delta > 0$ such that for every $x\in (-\delta, \delta)$ follows $|\sin(x)| < \sqrt{\epsilon}$. Now, by Dirichlet's approximation theorem, there exist integers $k$ and $n$ with 
$$ |2\pi k - n| < \delta. $$
Finally, by periodicity we have
$$ \sin(n)^2 = \sin(n - 2\pi k)^2 < \epsilon. $$
