How are these two definitions of the function $\sin x$ related? The high school definition of the sine function can be summarized by the following picture

This Wikipedia article says that the sine function can be defined by series. 
How are these two definitions related? (Since they both define the same thing, I think they should be "equivalent". But I don't see why.)
My second question might be vague: in practice, are there any concerns about which definition one uses?
 A: The geometric definition of sine is not entirely satisfying from a rigorous point of view, because it relies on the concept of angle and on geometry. Nothing really bad, but a purely analytic definition seems preferable, so to make analysis self-contained.
If the geometric definition is analytically sound, then the Taylor series for the sine converges on the whole real line. So we can directly define the function using the series it should expand to.
This has several advantages, for example no appeal to geometric intuition for computing the derivative (the well known argument for $\lim_{x\to0}\frac{\sin x}{x}=1$, that uses area or arc length); another advantage is the straightforward extension of the sine to complex numbers so to get in full generality the Euler relation
$$
e^{iz}=\cos z+i\sin z
$$
just by algebraic manipulation of (absolutely) convergent series.
A: The definition of $sin(x)$ function is indeed the same as in your image. 
To understand the series expansion of $sin(x)$, you need calculus. In case you do not have calculus yet. You may think it as an approximation by polynomials. 
For instance 
$sin(\frac{\pi}{6})=\frac{1}{2}=0.5$.
Let us just use two terms $x-\frac{x^3}{3!}$, where $3!=3*2*1=6$. Take $\pi\approx 3.1415$, then $\frac{\pi}{6}\approx0.5236$, then $x-\frac{x^3}{6}\approx=0.5236-\frac{(0.5236)^3}{6}=0.4997$, which is very close to $0.5$. For a general number $x$, you probably need more terms to get a better approximation. If you use all terms, it turns out equal. You can do some computation using known sin values to have a feeling.
A: Another point:  If we define $sin(x)= x- x^3/3!+ x^5/5!- x^7/7!+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$ then, differentiating term by term, $(sin(x))'= 1- x^2/2!+ x^4/4!- x^6/6!+ \cdot\cdot\cdot+ x^{n-1}/(n-1)!+ \cdot\cdot\cdot$ and $(sin(x))'= -x+ x^3/3- x^5/5!+ \cdot\cdot\cdot+ x^{n-2}/(n-2)!= - sin(x)$.  That is y= sin(x), defined as that series, satisfies the differential equation [tex]y''+ y= 0[/tex], with the initial condition y(0)= 0.  And, it is shown in Calculus that y= sin(x), with the usual definition also satisfies that equation and condition.
