Equivalence of $\pi$ is the first positive zero of the taylor series for $\sin(x)$ and $\pi/4 = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$ For $x\in\mathbb{R}$, define $\sin (x) = x - x^3/3!+x^5/5!-\cdots$ and $\pi = 4(1-\frac{1}{3}+\frac{1}{5} -\frac{1}{7}+\cdots)$. Then show that $\sin(\pi/2) = 1$
In the prologue of Real and Complex Analysis by Walter Rudin, pi is defined as the first positive zero of the series defined as $\sin(x)$; I want to check that pi is same as above defined Pi.

Here's a detailed outline for how to do this in a fairly standard line of development of trig functions from their power series definitions.
(A "plug in" combinatorial proof would be neat to see as well, of course.)
Step 1. Show the basic trig identities (double angle formulas and $\sin(x)^2 + \cos(x)^2 = 1$) for the taylor series definitions of $\sin(x)$ and $\cos(x)$.
[The double angle formulas are nice to prove using the power series for $e^x$. The second is immediate upon differentiation.]
Step 2. Let $pi$ (not the greek symbol for now) be the first root of $\sin(x)$. From the derivative of $\sin(x)$ at zero and the intermediate value theorem, we get that $\sin(x) > 0$ for $0 < x < pi$.
Step 3. Use trig identities to show that $\sin(2x) = 0$ if and only if $\sin(x) = 0$ or $\cos(x) = 0$. Conclude that $x = pi/2$ is the first time $\cos(x) = 0$.
Step 4. Let $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Show that $\frac{d}{dx} \tan(x) = \frac{1}{\cos(x)^2}$. Conclude that $\tan(x)$ is well-defined, increasing, and unbounded on $(-pi/2,pi/2)$.
Step 5. Define $\arctan(x)$ to be the inverse of $\tan(x): (-pi/2,pi/2) \rightarrow \mathbb{R}$. Show that $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$ from trig identities.
Step 6. Show that $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ from the series expansion for $\frac{1}{1+x^2}$ and differentiation of power series theorems and the fact that $\tan(0) = 0$. This converges and hence is valid on $(-1,1)$.
Step 7. The series also converges for $x = 1$ and hence (by a theorem on power series convergence), we get $\arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$. Thus it remains to show $2 \arctan(1) = pi/2$.
Step 8. Show from trig identities that $\tan(x) = \pm 1$ if and only if $\sin(x) = \pm \cos(x)$ if and only if $\cos(2x) = 0$. Conclude that $2 \arctan(1) = pi/2$, as desired (because $2 \arctan(1)$ must be the first zero of $\cos(x)$).
