Does the expansion of $\sin x$ contradict the normal formula $\sin x = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$? Lets say I have a right angled triangle with sides $3, 4$ and $5$ units. They form a perfect Pythagorean triplet. One of the angles in the triangle, say $\alpha$ must have the following condition:
$$\sin\alpha=\frac35$$
On the other hand, the expansion of the $\sin$ function shall never give such a definite answer.
Do they contradict? Or am I missing something?
If they contradict, which one is correct (more correct)?
 A: No, an infinite series will sum exactly to the value you are looking for, when you put in the $\alpha=\arcsin\frac35$ value, which is not a very nice number. It works in the forward direction ($\sin\alpha$ from $\alpha$, not the other way around). For the inverse function, use an expansion of $\arcsin x$.
You are not actually computing $\sin\alpha$ from $\alpha$. You are just rescaling the value that already is propotional to $\sin\alpha$ by definition. You have the sine, but you have no idea what the angle is, so calculation of $\sin\alpha$ was never actually performed.

$\sin\alpha$ is a function that takes a number $\alpha$ and gives you back $\sin\alpha$. The infinite series that you found is one of the ways to do that.
You don't actually know $\alpha$ in your triangle. You just marked the angle and figured out from the triangle's geometry what $\sin\alpha$ is. This means you never actually computed the function by giving it a known value.
In short - now you know $\sin\alpha$ but don't know $\alpha$. You could get $\alpha$ from this, if you used the inverse function of $\sin$: $\alpha=\arcsin\frac{3}{5}$. So your problem is not how to calculate the $\sin\alpha$ from $\alpha$ but exactly the inverse problem: how to get $\alpha$ from $\sin\alpha$. And that's a different thing.
You can compute the function $\alpha=\arcsin x$ with a series too. For instance, check out the infinite series here:
http://en.wikipedia.org/wiki/Arcsin#Infinite_series
To summarize: computing the function $\sin\alpha$ for a general value of $\alpha$ is not something you can do with elementary functions (exceptions are the "nice" angles with known values). Otherwise you wouldn't need a calculator for that! You can compute the values using infinite series, or some other technique. Your division didn't actually compute the function, it just told you the value for some angle.
Note that once you have $\sin\alpha$, you can get other trigonometric functions of the same angle easily from formulas like $\cos\alpha=\sqrt{1-\sin^2\alpha}$ and $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$. But that's another story...
