Factorising trigonometric functions In order to factorise $x^2-1$ one way of thinking about it would be to set it equal to zero and solve to get $x=1$ and $x=-1$.
You can then write $x^2-1=(x+1)(x-1)$
Can we do the same with trigonmetric functions, i.e
sin$x=0 \implies x=n\pi$ so 
sin$x=x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)...$
 A: Your idea is good.  It is well known that the sine function can be written as
$$\sin \pi x =\pi x \prod_{n=1}^{\infty}(1-x^2/n^2)$$
A: No. Try $x=\frac{1}{2}\pi$ and see that it doesn't work. 
A: Note that actually given the roots of a quadratic, you don't have a full specification. The same occurs with $\sin$ as well. Note that $\sin(x)$ and $2\sin(x)$ have the same roots for example. Also you must be very careful with infinite products such as the one you just wrote down.
A: You must make sure that ratio of both sides is a bounded function. Since that ratio only has removable singularities, it is a complex analytic function, the boundedness then implies by Liouville's theorem that the ratio is actually a constant.
A: Sort of. The product you are suggesting does not converge except at point $x=n\pi$. But a variant does work.
The function $\sin(\pi z)$ can be expressed as the infinite product
$$\pi z\prod_1^\infty \left(1-\frac{z^2}{n^2}\right).$$
From this you can get an infinite product representation for $\sin x$ by setting $z=\frac{x}{\pi}$.
This result is due to Euler, and played a crucial part in his famous  proof of the fact that $1+\frac{1}{ 2^2}+\frac{1}{3^2}+\cdots =\frac{\pi^2}{6}$. 
