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?
The answers so far (which essentially say: "sine is not a polynomial") are ok. However, they fail on the heuristic side. Thinking like this Euler would have never solved the Basel problem.
Here is my attempt:
There are two main problems with your answer:
1) sin only has a single root at $x=0$ whereas your expression has a double root
2) one can expand a polynomial in terms of factors however there is an overall factor that has to be adjusted
To 2): we have that $4x^2-1$ has the solutions $x=\pm 1/2$. However, it is of course not true that $4x^2-1 = (x-1/2)(x+1/2)$ but rather $4x^2-1 =\mathbf{4} (x-1/2)(x+1/2)$
So let us try to fix the problems and propose
$$\sin x = \lim_{N\to \infty} c x \prod_{n=1}^N (x^2-n^2\pi^2) \tag{1}$$ with $c$ a constant that has to be still determined. (It will turn out that the constant $c$ has to depend on $N$ and that $c_N \to \infty$ for $N\to\infty$)
Now, how to set the overall constant $c$: The simplest way is to remember that $\sin x \sim x$ for $x\to0$. Looking at small $x$, we immediately recognize that $$c = c_N = \prod_{n=1}^N ({-n^2\pi^2})^{-1}$$ is the proper normalization factor. With that, we obtain $$\sin x = x \lim_{N\to\infty} \prod_{n=1}^N\left(\frac{x^2-n^2 \pi^2}{-n^2 \pi^2}\right) = x\prod_{n=1}^\infty \left(1- \frac{x^2}{n^2 \pi^2}\right). $$
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 $x\in\mathbb R$.
