How do I prove that $3<\pi<4$?

Let's not invoke the polynomial expansion of $\arctan$ function.

I remember I saw somewhere here a very simple proof showing that $3<\pi<4$ but I don't remember where I saw it.. (I remember that this proof is also in Wikipedia)

How do I prove this inequality?

My definition for $\pi$ is the twice the first positive real number such that $\cos x= 0$ where $\cos x = \frac{e^{ix} + e^{-ix}}{2}$

After all the comments it's unclear what is allowed and what is forbidden. I'd suggest the following: $${1\over 2}=\sin{\pi\over 6}<{\pi\over 6}<\tan{\pi\over 6}={1\over\sqrt{3}}\ ,$$ which leads to $3<\pi<3.5$.

Starting from the antique definition of $\pi$ : The length of cirle of radius $1$ is $=2\pi$.

On the figure below, obviously the length of the circle is between $6$ and $8$ (lengths of the hexagon and square, respectively)

So, $6 < 2\pi < 8$ hence $3 < \pi < 4$