I wonder if the graphs would both have the same $x$ intercepts (due to the zero product property) and continue to infinity. Would this also mean that $y=\prod_{n=0}^\infty (x^2-[n\pi]^2)$ equals $y=\sin(x)$ due to difference of two squares?
Edit: Fixed math markup and slight math error.
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Your function doesn't exist when $x\ne k\pi$ $\forall k\in\mathbb{Z}$. First, let's simplify the productand (for lack of a better word): $$f(x)=\prod_{n=0}^\infty(x^2-(n\pi)^2)=x^2(x^2-\pi^2)(x^2-4\pi^2)...$$Let $x^2=\epsilon$ and $(k+1)^2\pi^2>\epsilon>k^2\pi^2$. Then we have $$\epsilon(\epsilon-\pi^2)(\epsilon-4\pi^2)...(\epsilon-(k\pi)^2)(\epsilon-(k+1)^2\pi^2)...$$The left part is positive and the right part is negative. They all have a magnitude greater than $1$ except the expression in between the three dots. This makes it non-convergent except when $x$ is an integer multiple of $\pi$.
The difference of two squares should actually read $x^2-n^2\pi^2$. Your presumption is purely heuristic but it can be made to be true, and this is what Euler famously did. Observe your product, as written, never converges. By some miracles it turns out the following adjustment is correct: $$\forall x\in\Bbb R:\quad\quad\sin x=x\cdot\prod_{n\ge1}\left(1-\frac{x^2}{\pi^2n^2}\right)$$Called the Weierstrass or Euler product formula for the sine function. This applies difference of two squares to the root factors $(1-x/\pi n)(1+x/\pi n)$.
However it is not super trivial to prove this, and requires some kind of careful real or complex analysis. It’s not enough to just say: “look! They have the same zeroes”. There is a related general theorem which apparently allows you to do something like this but care is still required.