<s>[INST] How to transform the factored form of $\sin(x)$? We know $\sin(x)=0$ has solutions $0,\pm\pi,\pm2\pi,\pm3\pi,\dots$.
So $\sin(x)$, if interpreted as a polynomial, could be written as:
$a_0x^0+a_1x^1+a_2x^2+\cdots$ and we know this polynomial too:
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$
So, the question is, is it possible to transform the factored form of $\sin(x)$:
$$\sin(x)=a x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\dots$$
to
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\ ?$$
 [/INST] The proposal is:
$$\sin x = a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots$$
The standard result, already posted by Sam, is in effect
$$
\sin x = \frac{x-\pi}{\pi}\cdot \frac{x+\pi}{\pi} \cdot \frac{x - 2\pi}{2\pi}\cdot \frac{x+2\pi}{2\pi}\cdot\frac{x-3\pi}{3\pi}\cdot\frac{x+3\pi}{3\pi} \cdots
$$
So the coefficient "$a$" in front of the whole thing is more . . . . interesting . . . than might be initally guessed.  Might Euler have considered
$$
a = \frac{1}{\pi^2}\cdot\frac{1}{(2\pi)^2}\cdot\frac{1}{(3\pi)^2}\cdots
$$
to be some sort of "infinitely small number"?  Might it actually be fruitful in some way to think of it that way?
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