I was reading Euler's proof that $\sum_n n^{-2}=\frac{\pi^2}{6}$ and I don't agree with his reasoning. My issue is outlined below.
First, Euler observed that the function $\sin x$ has roots at $x = 0, ±π, ±2π, ±3π, \ldots$. Next, he observed that the infinite product $x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2})\ldots$ also has roots at $x = 0, ±π, ±2π, ±3π, \ldots$.
Euler then believed that these two functions are equivalent.
I have a problem with the above statement. Obviously, having the same roots does not imply that the two functions are equivalent. For instance, $f(x)=0$ also has roots at $x = 0, ±π, ±2π, ±3π, \ldots$.
Euler then goes on to use the infinite product and Maclaurin expansion of $\sin x$ to compare coefficients of $x^3$. But his belief that those two functions are equivalent stains all subsequent parts of the proof.
Am I missing something obvious here? Or was Euler not rigourous enough in his proof?
Your example involving the null function makes no sense. Euler did not just claim that all the zeros of the sine function are also zeros of$$x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\left(1-\frac{x^2}{(3\pi)^2}\right)\ldots;\tag1$$he claimed also that both functions have the same set of zeros.
But, yes, is argument is no rigorous by modern standars. How did he know that the sine function had no other (complex non-real) zeros besides those from $\pi\mathbb Z$? I suppose that that's way he extended the sine function to $\mathbb C$ (and, no, there are no more zeros).
And of course, even assuming that, one still has to do something else to justify the equality $\sin x=(1)$.
You will find here a list of proofs of that equality, compiled by our former moderator Robin Chapman.