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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 $3 < \pi < 4$
I assume that $x\to e^x$ is the unique differentiable function that is its own derivative and maps $0\mapsto 1$. From this (especially using uniqueness) one quickly establishes $e^{x+y}=e^xe^y$, $e^{\bar z}=\overline{e^z}$ etc.
With the definitions $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ we then see that $\cos $ and $\sin$ map reals to reals and that $\cos^2x+\sin^2x=1$, $\sin'=\cos$, $\cos'=-\sin$ etc.
Claim 1: There is no $a$ such that $\sin x>\frac12$ for all $x\in [a,a+4]$.
Proof: Otherwise, $2\ge\cos (a+2)-\cos a=\int_a^{a+4}\sin x\,\mathrm dx>2$. $_\square$
Claim 2: There exists $a>0$ such that $\cos a=0$.
Proof: Otherwise we would have $\cos x>0$ for all $x>0$, hence $\sin x$ strictly increasing towards some limit $s:=\lim_{x\to+\infty}\sin x\in(0,1]$. Then $\lim_{x\to\infty}\cos x=\sqrt{1-s^2}=0$ and hence $s=1$. On the other hand, $s\le \frac12$ by claim 1. $_\square$.
By continuity, there exists a minimal positive real $p$ such that $\cos p=0$.
By the IVT, $-\sin \xi =\frac{0-1}{p}$, i.e. $\sin\xi=\frac1p$ for some $\xi\in(0,p)$. This already shows $$p\ge1.$$ Consider the function $f(x)=\cos x-1+\frac12x^2$. We have $f'(x)=-\sin x+x$, $f''(x)=1-\cos x\ge 0$ for all $x$; hence $f'(x)\ge f'(0)$ for all $x\ge 0$, i.e. $\sin x\le x$ for all $x\ge 0$; hence $f(x)\ge f(0)=0$ for all $x\ge0$. We conclude $\cos 1\ge \frac12$. In fact the iniequality is strict, i.e. $\cos1>\frac12$ as otherwise we'd have $f''(x)=0$ for all $x\in[0,1]$. From $\cos'=-\sin\ge -1$ we then see that $\cos (1+x)>\frac12-x$ for $x> 0$. Thus $$p> \frac32.$$
Consider $g(x)=\cos x-1+\frac12x^2-\frac1{24}x^4$. Then $g'(x)=-\sin x+x-\frac16x^3$, $g''(x)=-\cos x-1-\frac12x^2=-f(x)\le 0$ for $x\ge 0$. Hence $g'(x)\le g'(0)=0$ for all $x\ge 0$ and finally $g(x)\le g(0)=0$ for all $x\ge 0$. We conclude $0\ge g(2)=\cos 2-1+\frac42-\frac{16}{24}$, i.e. $\cos 2\le-\frac13$ and hence $$ p<2.$$
We need to show that $3/2<\pi/2<2$, so (using your definition of $\pi$) that $\cos 2<0$ and $\cos t>0$ for all $0\leq t\leq 3/2$.
The Taylor series for $\cos t$ follows immediately from the series for $e^t$ and your definition of $\cos t$. The signs alternate. The terms decrease in absolute value for $t=2$ (except for the second term), so taking the first three terms give the upper bound $\cos 2 < -\tfrac{1}{3}$.
On the other hand for $t=\tfrac{3}{2}$, the first four terms give the lower bound $\cos \tfrac{3}{2}>0.0701171875$.
Finally, $\cos' = -\sin$ (immediate from your definitions), so if $\sin t\geq 0$ for $0\leq t\leq 3/2$ we will know that $\cos$ is decreasing in that interval and so cannot be negative. For this interval, $\sin t > t-t^3/6$ (same argument about alternating terms with decreasing absolute value); $t-t^3/6$ is positive between its zeros at $t=0$ and $t=\sqrt{6}>3/2$.
Well you can take that $$\cos 6>0\\\cos 8<0$$ Since $\cos$ is a continuos function,a zero($\pi$) has to be between $6$ and $8$.If you have to prove that $\cos 6> 0 \land \cos 8<0$,I guess you could do so with Taylor series of $e^{ix}+e^{-ix}$
Similarly to
$$ \frac{22}{7}-\pi=\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx $$ Why do we need an integral to prove that $\frac{22}{7} > \pi$?
we have $$ 4-\pi=4\int_0^1\frac{x^2}{1+x^2}dx $$
and $$ \pi-3=2\int_0^1\frac{x(1-x)^2}{1+x^2}dx $$
https://math.stackexchange.com/a/1593090/134791
Combining them $$3+2\int_0^1\frac{x(1-x)^2}{1+x^2}dx=\pi=4-4\int_0^1\frac{x^2}{1+x^2}dx$$ and noting that the integrands are positive for $0<x<1$ $$3<\pi<4$$
