How to transform the factored form of $\sin(x)$? We know $\sin(x)=0$ has solutions $0,\pm\pi,\pm2\pi,\pm3\pi,\dots$.
So $\sin(x)$, if interpreted as a polynomial, could be written as:
$a_0x^0+a_1x^1+a_2x^2+\cdots$ and we know this polynomial too:
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$
So, the question is, is it possible to transform the factored form of $\sin(x)$:
$$\sin(x)=a x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\dots$$
to
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\ ?$$
 A: The proposal is:
$$\sin x = a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots$$
The standard result, already posted by Sam, is in effect
$$
\sin x = \frac{x-\pi}{\pi}\cdot \frac{x+\pi}{\pi} \cdot \frac{x - 2\pi}{2\pi}\cdot \frac{x+2\pi}{2\pi}\cdot\frac{x-3\pi}{3\pi}\cdot\frac{x+3\pi}{3\pi} \cdots
$$
So the coefficient "$a$" in front of the whole thing is more . . . . interesting . . . than might be initally guessed.  Might Euler have considered
$$
a = \frac{1}{\pi^2}\cdot\frac{1}{(2\pi)^2}\cdot\frac{1}{(3\pi)^2}\cdots
$$
to be some sort of "infinitely small number"?  Might it actually be fruitful in some way to think of it that way?
A: The answer is yes, you can factor $\sin(z)$ into a product of zeros. The general theory behind this is Weierstrass factorization. For your example,
$$\sin(z)=z\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2\pi^2}\right)$$
In fact, Euler famously used an unrigorously derived form of this identity to solve the Basel problem. I say "unrigorously" here because, while one can show that the function can be written as a product over it's zeros, it's the outside term (1 in this case, infront of the first 'z') that takes work to derive. For example, if we had the function $e^z\sin(z)$, then there would be an $e^z$ factor on the outside of the product. Since $e^z$ doesn't have any zeros, you cannot break it down into such a product, so you just tack it on as a factor. More difficult functions have even more intricate product representations but the general rule of thumb is that the function factors into a product over zeroes times something that looks like $e^{g(s)}$. 
An interesting consequence of this is that it's not necessarly possible to directly transform such a product into the infinite polynomial corresponding to the function. One can however, write down a correspondence between products and sums of the zeroes and the polynomial coefficients. This comes from Vieta's formulas which is precisely what Euler used to show $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$. 
