How does analytic continuation lets us extend functions to the complex plane? I'm trying to understand analytic continuation and I noticed on wolfram that it  

allows the natural extension of the definition trigonometric,
  exponential, logarithmic, power, and hyperbolic functions from the
  real line $\mathbb{R}$ to the entire complex plane $\mathbb{C}$

So how does it extend, say, $f(x) = \sin(x)$, $x \in \mathbb{R}$ to the complex plane? What are the steps that have to be taken to extend this function (and others) to the complex plane?
 A: There are various ways to extend a function $f:\>J\to{\mathbb R}$ defined on some open interval $J\subset{\mathbb R}$ to a holomorphic function $\tilde f:\>\Omega\to{\mathbb C}$, if such an extension is at all possible.  But whatever method you use, you will always end up with the same function $\tilde f$.
(i) If $f$ is given as a rational expression $f(x)={p(x)\over q(x)}$ with $p$ and $q$ polynomials, $q(x)\ne0$ on $J$, then the extension is simply
$$\tilde f(z):={p(z)\over q(z)}\ ,$$
and is defined on all of ${\mathbb C}$, minus the complex zeros of $q$.
(ii) If $f$ is given by some power series with positive radius of convergence $\rho$: $$f(x)=\sum_{k=0}^\infty c_k(x-a)^k\qquad(|x-a|<\rho)$$
then you may again put
$$\tilde f(z):=\sum_{k=0}^\infty c_k(z-a)^k\qquad(|z-a|<\rho)$$
and can be sure that $\tilde f$ is holomorphic at least in the disk $D_\rho(a)$.
(iii) If $f$ is the solution of some IVP $y'=g(x,y)$, $y(x_0)=y_0$,  and $g$ is "real analytic" (example: $y'=\lambda y$) then this ODE can be viewed at also in a complex sense (i.e., $x$ and $y$ are complex variables). The IVP then defines a holomorphic function $\tilde f$ in the neighborhood of $(x_0,y_0)$, which is the extension of $f$ to the complex domain.
An example: Euler's formula gives $$\sin t={e^{it}-e^{-it}\over 2i}\ .$$ We can then say right away that
$$\sin z:={e^{iz}-e^{-iz}\over 2i}$$
is the complex extension of $\sin$.
A: More interesting is the case that Riemann worked out:
$$\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^s}{e^x-1}\frac{dx}{x},$$
where $\Gamma(s)$ is the gamma function, $\zeta(s)$ is the zeta function and $s\in\mathbb{C}.$ To extend this formula to $\mathbb{C},$ Riemann considers the path integral, on the complex plane, 
$$\oint_C\frac{(-z)^s}{e^z-1}\frac{dz}{z},$$
where the path $C$ goes from $\infty$ to the origin $O$, above the positive real axis, and circulating $O$, counterclockwise through a circumference of radius $\delta$, say, returning to $\infty$ along the bottom of the positive real axis. The important thing, for the evaluation of above integral, is that we may split it into three integrals, namely
$$\biggl(\int_\infty^\delta+\int_{|z|=\delta}+\int_\delta^\infty\biggr)\frac{(-z)^s}{e^z-1}\frac{dz}{z},$$
recalling that $(-z)^s=e^{s\log(-z)}$, and $\log(-z)=\log|z|+ i\, \text{arg}(-z).$ So that, in the first integral, when $-z$ lies on the negative real axis, we take $\text{arg}(-z)=-\pi\,i;$ on the second one $-z=-\delta,$ and as $-z$ progress counterclockwise about $O$, $\text{arg}(-z)$ goes from $-\pi\,i$ to $\pi\,i.$ Finally, in the last integral $\text{arg}(-z)=\pi\,i,$ therefore the first and third integrals do not cancel. The second integral cancels as $\delta\to 0.$ The rest is purely technical and leads to the analytical continuation of the $\zeta(s)$ function of Riemann everywhere except 1 where it has a simple pole with residue 1.
