Euler Formula for Sine Function/Real Numbers/Proof 3

Proof
We have that $\sin x$ has a power series representation:


 * $\sin x = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots$

The roots of sine are the numbers $k \pi$, where $k$ is any integer.

From the Polynomial Factor Theorem, the following might be true:


 * $\ds \sin x = A x \prod \paren {1 - \frac x {k \pi} }$

where the product is taken over all $n \in \Z \setminus \set 0$, and $A$ is some constant.

The intuition is as follows.

From Limit of $\dfrac {\sin x} x$ at Zero:
 * $\dfrac {\sin x} x \to 1$ as $x \to 0$

Letting $x$ tend to $0$ in the above equation implies that $A = 1$.

We now formalize the above claims.