Why can't we write $\sin x$ as $\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$? Why can't we write $\sin x$ as $\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$?
Since $ n\pi$ where $n \in \mathbb{N}$ are all the roots of $\sin{x}$, then by the fundamental theorem of arithmetic it may be written as
$$\sin x=\prod_{n=0}^{\infty}\left(x^2-n^2\pi^2\right)$$
But this does not actually represent $\sin{x}$ as shown by graph below

Then what modifications have to be made to make both graphs look similar, Please Explain?

Because, apart from the fact that $\sin x$ is not a polynomial (the fundamental theorem of Algebra refers to polynomials) the infinite product
$$
\prod_{n=0}^\infty(x^2-n^2\pi^2),
$$
does NOT converge, for any $x\in\mathbb R$.
