Trigonometric Coincidence I Know that using Taylor Series, the formula of $\sin x$ is
$$x-x^3/3!+x^5/5!-x^7/7!\cdots,$$
and the unit of $x$ is radian (where $\pi/2$ is right angle).
However, the ratio of the circumference and the diameter of a circle is also $\pi$. Is it a coincidence? Or is there a proof?

The facts you mention are definitely not a coincidence; they are a consequence of how things are defined, and partly a reason why they are defined so. (This is an incredibly vague statement, but the question itself is a little vague).
Some textbooks will define $\sin x$ by the series expansions you just mentioned, and then define $\pi$ as the least such positive number that $\sin \pi = 0$. Also, put $\cos x = \sin' x$. You can then check that $(\cos x, \sin x)$ parametrize points on the circle, and then calculate the circumference of a unit circle using the appropriate integral. This is a legitimate way of doing mathematics, but not a very illuminating one (in this particular situation), I'm afraid. 
If you know a little about differential equations (and are prepared not to be absolutely rigorous), you can proceed as follows. Let's agree that you know that there are functions $\sin x $ and $\cos x$ (as functions of the angle, so that the precise units are yet to be determined). Denote the measure of the full angle by $\tau$ (there $\tau$ is just a name just now, nothing more), and the circumference of the unit circle by $C$. Note that we could set $\tau$ to be pretty much any number (like the well know constant $2 \pi$, or $360$, or whatever), but $C$ is fixed once and for all, and in fact $C = 2 \pi$.
Now, $\sin x$ and $\cos x$ will have some expansions as Taylor series, which will depend on what choice of $\tau$ we take (e.g. you have $\sin \tau/4 = 1$, but what this means depends on what $\tau$ is). We would like to choose $\tau$ so that the Taylor series for $\sin$ is something ``nice'' - which is of course a very vague requirement. To figure out the Taylor series, it's easiest to start with some differential equations. So let's try to find out what $\sin'x$ is. After some thought (it's probably best to draw a picture...) you can convince yourself that $\sin'x = \frac{C}{\tau} \cos x$. Likewise,  $\cos'x = - \frac{C}{\tau} \sin x$. Hopefully, it is clear that things will become nicest when $C = \tau$, so that $\sin' x = \cos x $ and $\cos'x = - \sin x$ - and this is the choice that the mathematicians made! 
With a little work, you can derive from the conditions $\sin' x = \cos x $ and $\cos'x = - \sin x$ the Taylor expansion for $\sin x $ and $\cos x$. You could set other units (i.e. choose different $\tau$) - only then the Taylor expansion would not look so nice.
