How to show that $\tan(n), n\in \mathbb{N}$ is not bounded

Solution 1:

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 $\tan(n)$.

Solution 2:

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.