Understanding the derivation of Euler's Basel formula.

Question

I was looking at the derivation of Euler's Basel formula and I had a question.

Since the zeroes of $\sin(x)$ are $0, ±π, ±2π,...$ , we can factorise $\sin (x)$ :

$$\sin(x) = ax\cdot(x-π)\cdot(x+π)\cdot(x-2π)\cdot(x+2π)\cdots$$

and the place where I found this equation gives the next simplification as : $$\sin(x) = x\cdot(1-x/π)\cdot(1+x/π)\cdot(1-x/2π)\cdot(1+x/2π)\cdots$$ Since the limit as $x$ tends to $0$ of $\sin (x)/x = 1$

In the above equation shouldn't it be $x/π - 1$ and so on for the other such expressions and not the other way around and also how do you arrive at that expression? Is it by dividing each term of the factor with constant in it like $π, 2π$? but would not that change the value of the expression?

Answer 1

I have a bunch of links which should clarify things (mainly the first one):

How was Euler able to create an infinite product for sinc by using its roots?

https://en.wikipedia.org/wiki/Basel_problem

How do we know that the Sine function has no Non-Real Roots?

Answer 2

Note that since $$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots,$$ Euler concluded that the product of the roots of $\frac{\sin x}{x}$ is $1.$ Therefore $a$ in your expansion should be equal to $$\frac{1}{\prod_{k} (kπ)^2}.$$