Since the zeroes of $\sin x$ are $0, ±π, ±2π,...$ , we can factorise sin $x$ :
$$\sin(x) = ax.(x-π).(x+π).(x-2π).(x+2π)...$$
I would be really grateful if someone could help me understand the $a$ in the expression. and the place where I found this equation gives the next simplification as : $$\sin(x) = x\cdot(1-x/π)\cdot(1+x/π)\cdot (1-x/2π)\cdot(1+x/2π)\dotsb$$ 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?
