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 $3<\pi<3.5$.
