How to show that $\tan(n), n\in \mathbb{N}$ is not bounded I'm struggling on how to show that the sequence of $\tan(n)$ is not bounded. Can you please give me some help?
 A: Suppose $p,q$ are integers that are relatively prime, and assume that $p$ is odd. Then
$$\frac{nq-pk}{q}-\frac{p}{2q}=\frac{2nq-(2k+1)p}{2q}$$
Since $p$ and $2q$ are relatively prime, there exist $n>0$ and $k>0$ such that $2nq-(2k+1)p=1$. Thus there exists a value of $n$ such that $n$ mod $\frac{p}{q}$ is within $\frac{1}{2q}$ of $\frac{p}{2q}$. Now take $\frac{p}{q}$ to be an approximation of $\pi$ with a large denominator; then with $q$ large enough we can make $n$ mod $\frac{p}{q}$ arbitrarily close to $\frac{p}{2q}$, hence we can make $n$ mod $\pi$ arbitrarily close to $\frac{\pi}{2}$ for some value of $n$, resulting in arbitrarily large values of $\tan(n)$.
A: Hint. Prove that $\forall \varepsilon > 0$ there are infinitely many $n \in \mathbb{N}$ such that 
$$
n\ (\text{mod }2\pi) \in \left(\frac{\pi}{2}-\varepsilon, \frac{\pi}{2}+\varepsilon\right).
$$
The simplest way to do this is to use Dirichlet's principle.
A: Here you will find a simple proof of the fact that if $\alpha$ is any irrational number, $\{n\alpha\bmod1:n\in\Bbb Z\}$ is dense in $[0,1)$, where $x\bmod1=x-\lfloor x\rfloor$, the fractional part of $x$. (This is a useful fact, one worth knowing.) The argument still works if you replace $\Bbb Z$ by $\Bbb Z^+$.
In particular, we can take $\alpha=\frac1{\pi}$ to find that
$$\left\{\frac{n}{\pi}\bmod 1:n\in\Bbb Z^+\right\}$$
is dense in $[0,1)$. This gives us integers $n$ such that $\frac{n}{\pi}\bmod1$ is as close as we wish to $\frac12$, meaning that $n$ is as close as we want to a number of the form $m+\frac12$ with $m\in\Bbb Z$. We’ll scale it up by a factor of $\pi$ to get integers $n$ arbitrarily close to numbers of the form $\left(m+\frac12\right)\pi$ with $m\in\Bbb Z$, which will prove the result. (Why?)
For each $\epsilon>0$ there is an $n_\epsilon\in\Bbb Z^+$ such that 
$$\left|\left(\frac{n_\epsilon}{\pi}\bmod1\right)-\frac12\right|<\epsilon$$
and hence
$$\left|\pi\left(\frac{n_\epsilon}{\pi}\bmod1\right)-\frac{\pi}2\right|<\pi\epsilon\;.$$
Now
$$\pi\left(\frac{n_\epsilon}{\pi}\bmod1\right)=\pi\left(\frac{n_\epsilon}{\pi}-\left\lfloor\frac{n_\epsilon}{\pi}\right\rfloor\right)=n_\epsilon-\pi\left\lfloor\frac{n_\epsilon}{\pi}\right\rfloor\;,$$
so
$$\left|n_\epsilon-\pi\left(\left\lfloor\frac{n_\epsilon}{\pi}\right\rfloor+\frac12\right)\right|<\pi\epsilon\;.$$
Plainly we can make $\pi\epsilon$ as small as we like by choosing $\epsilon$ small enough, and $\left\lfloor\frac{n_\epsilon}{\pi}\right\rfloor+\frac12$ is of the form $m+\frac12$ for some integer $m$. Thus, we can find integers arbitrarily close to numbers of the form $\left(m+\frac12\right)\pi$, where $m$ is an integer, and the result follows.
