How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by only manipulating the series?

A fairly easy way to introduce $\pi$ in trigonometric functions defined by series is:


*

*Define


$$e^z=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$
Then use Euler's formula to define $\sin$ and $\cos$:
$$\sin z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$
$$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}$$


*Then prove that $\cos$ has a least positive root, which you call $\pi/2$.
For this, you can show easily that $\cos 0>0$ and $\cos 2<0$ (the latter using majoration of the rest in the series, which is alternating).

*Prove and use $e^{a+b}=e^ae^b$ (it's a Cauchy product) to derive similar identities for $\sin$ and $\cos$.

*Use (2) and (3) to prove that $\sin$ and $\cos$ are $2\pi$-periodic.

Here is the detailed derivation
First, define 
$$e^z=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$
The series converges for all $z\in\Bbb C$ by the ratio test, thus it defines an entire function on the complex plane. It is $C^{\infty}$ on $\Bbb C$, and the restriction to real $z$ is real-valued and also $C^\infty$.
Putting $z=0$, you have $e^0=1$, and by differentiating the series, you get $\dfrac{\mathrm{d}e^z}{\mathrm{d}z}=e^z$.
Then you define
$$\cos z=\frac{e^{iz}+e^{-iz}}{2}=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}$$
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$
And while we are at it,
$$\cosh z=\frac{e^{z}+e^{-z}}{2}=\sum_{n=0}^\infty\frac{z^{2n}}{(2n)!}$$
$$\sinh z=\frac{e^{z}-e^{-z}}{2}=\sum_{n=0}^\infty\frac{z^{2n+1}}{(2n+1)!}$$
Thus $\cos$ and $\cosh$ are even, while $\sin$ and $\sinh$ are odd.
Of course, the restriction of these functions to real $z$ are real-valued.
You have
$$e^{iz}=\cos z+i\sin z$$
And the derivatives $\sin'=\cos$ and $\cos'=-\sin$.
Notice also that the terms in the series of $e^x$ are increasing for increasing $x\geq0$, thus $x\rightarrow e^x$ is increasing for $x\geq0$ and you have $e^x\geq1+x$ for $x\geq0$, and $e^x\underset {x\rightarrow+\infty}\longrightarrow+\infty$.
Let $(a,b)\in\Bbb C^2$. Since the series of $e^z$ is absolutely convergent for all $z$, the following equality holds
$$e^ae^b=\sum_{i=0}^\infty \frac{a^i}{i!}\sum_{j=0}^\infty \frac{a^j}{j!}=\sum_{n=0}^{\infty} u_n$$
With
$$u_n=\sum_{k=0}^n\frac{a^kb^{n-k}}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n {n\choose k}a^kb^{n-k}=\frac{(a+b)^n}{n!}$$
Thus $e^ae^b=e^{a+b}$ for all complex $a,b$.
Thus you have $e^ze^{-z}=1$, and $e^z$ is never zero.

Digression on the real exponential
Hence for real $x$, $e^x\neq0$, and since the function is $C^0$ (even $C^\infty$), its sign does not change, and $\forall x\in\Bbb R, e^x>0$.
Also, since $e^xe^{-x}=1$ and $e^x\underset {x\rightarrow+\infty}\longrightarrow+\infty$, you have $e^x\underset {x\rightarrow-\infty}\longrightarrow0$.
And since the derivative of $e^x$ is itself, the derivative is also always positive, and the exponential is increasing on $\Bbb R$.
You can conclude it's a bijection, and since $e^ae^b=e^{a+b}$ and $e^0=1$, this proves that the exponential is a group isomorphism between $(\Bbb R,+)$ and $(\Bbb R^\star_+,\cdot)$.
Call $\log$ the inverse isomorphism, defined on $\Bbb R^\star_+$, with $\log (ab)=\log(a)+\log(b)$ for all $a>0, b>0$. Also, using the formula of derivation of an inverse function, you have $\log'(x)=1/x$.

Trigonometric identities
From $e^ae^b=e^{a+b}$ and using Euler's identity, you can derive the usual trigonometric (and hyperbolic trigonometry) identities. I'll show how on an example:
$$\cos a\cos b-\sin a\sin b=\frac{e^{ia}+e^{-ia}}{2}\frac{e^{ib}+e^{-ib}}{2}-\frac{e^{ia}-e^{-ia}}{2i}\frac{e^{ib}-e^{-ib}}{2i}$$
$$=\frac14\left[(e^{ia}+e^{-ia})(e^{ib}+e^{-ib})+(e^{ia}-e^{-ia})(e^{ib}-e^{-ib})\right]$$
$$=\frac{1}{4}\left[\left(e^{i(a+b)}+e^{i(a-b)}+e^{i(b-a)}+e^{-i(a+b)}\right)+\left(e^{i(a+b)}-e^{i(a-b)}-e^{i(b-a)}+e^{-i(a+b)}\right)\right]$$
$$=\frac{e^{i(a+b)}+e^{-i(a+b)}}{2}=\cos(a+b)$$
Likewise, you have $\sin(a+b)=\sin a\cos b+\sin b\cos a$, and a bunch of other formulas.
In particular, you have for all $z\in\Bbb C$:
$$\cos^2 z + \sin^2 z=\cos(z-z)=1$$
$$\cos 2z=\cos^2 z-\sin^2 z=2\cos^2 z-1$$
These are true for real $z$, and since the functions are then real-valued, you have $|\cos x|\leq 1$ and $|\sin x| \leq 1$ for all $x\in\Bbb R$.

Definition of $\pi$
You have $\cos 0=1$ from the series definition, and
$$\cos 2=\sum_{n=0}^\infty (-1)^n\frac{2^{2n}}{(2n)!}$$
The series is alternating with decreasing term after $n=1$, thus
$$\cos 2<1-\frac{2^2}{2!}+\frac{2^4}{4!} = -\frac13 <0$$
Since $\cos$ is continuous, it has at least one root in $]0,2[$.
The series for $\sin x$ is also alternating for $0< x\leq 2$, and its general term is decreasing after $n=0$, thus for $x\in[0,2]$,
$$\sin x \geq x-\frac{x^3}{6}=x\left(1-\frac{x^2}{6}\right)$$
The RHS of the inequality has roots $0$ and $\pm\sqrt{6}$, and $\sqrt{6}>2$, thus for $x\in]0,2]$, $\sin x>0$.
Since $\cos'=-\sin$, you have that the function $\cos$ is decreasing on $]0,2[$.
Therefore, $\cos x=0$ has one and only one root in $[0,2]$. Let's call this root $\frac{\pi}2$.
We have then $\cos \frac{\pi}2=0$, thus $\cos^2 \frac{\pi}2+ \sin^2 \frac{\pi}2=1$ implies $\sin \frac{\pi}2=\pm1$, and since it's positive, $\sin \frac{\pi}2=1$.
Also, from $\cos 2x=2\cos^2x-1$, you get that $\cos \pi=-1$, and then $\sin\pi=0$.
Notice that you have also
$$e^{i\pi}=\cos \pi+i\sin\pi=-1$$

Trigonometric functions are periodic
From the identities
$$\cos (a+b)=\cos a\cos b - \sin a \sin b$$
$$\sin (a+b)=\sin a\cos b + \cos a \sin b$$
You get
$$\cos (a+\pi)=\cos a\cos \pi - \sin a\sin \pi=-\cos a$$
$$\sin (a+\pi)=\sin a\sin \pi + \cos a\sin \pi=-\sin a$$
And finally
$$\cos (a+2\pi)=\cos a$$
$$\sin (a+2\pi)=\sin a$$
Thus $\cos$ and $\sin$ are $2\pi$-periodic. We have still to prove it's the smallest possible period, but before, let's have a look at variations of $\cos$ and $\sin$ on one period $[0,2\pi]$.
We already know that for $x\in[0,\pi/2]$, $\cos x\geq 0$ and $\sin x\geq 0$, where the former is decreasing from $1$ to $0$, and the latter is increasing from $0$ to $1$.
First, we complete an half-period. Using the previous identities:
$$\cos (\pi-x)=-\cos x$$
$$\sin (\pi-x)=\sin x$$
Thus for $x \in [\pi/2,\pi]$, $\cos$ is decreasing from $0$ to $-1$, and $\sin x$ is decreasing from $1$ to $0$.
Then we complete the full period with
$$\cos (a+\pi)=-\cos a$$
$$\sin (a+\pi)=-\sin a$$
This means that for $x\in[2\pi]$, the only roots of $\cos x$ are $\pi/2$ and $3\pi/2$, and the only roots of $\sin x$ are $0$, $\pi$ and $2\pi$.
Now, is $2\pi$ the smallest period? Suppose there is a $\lambda \in ]0,2\pi[$ such that for all $a$, $\cos (a+\lambda)=\cos a$, then
$$\cos(a+\lambda)=\cos a\cos\lambda-\sin a\sin \lambda=\cos a$$
And for $a=\pi/2$,
$$-\sin \lambda=0$$
Thus $\lambda=\pi$, but then $\cos a=\cos(a+\lambda)=-\cos a$, which is not true for example for $a=0$. Thus $2\pi$ is the minimal period.

What next?
You could define $\tan x=\frac{\sin x}{\cos x}$ and derive identities, then define inverse trigonometric functions on some wise restriction (since a periodic function has no inverse), and also define $a^b=e^{b\log a}$. And you have a construction of all so-called elementary functions.
