Euler Formula for Sine Function

Theorem

 * $\displaystyle \frac {\sin x} x = \left({1 - \frac{x^2}{\pi^2}}\right) \left({1 - \frac{x^2}{4 \pi^2}}\right) \left({1 - \frac{x^2}{9 \pi^2}}\right) \cdots = \prod_{n \mathop = 1}^\infty \left({1 - \frac{x^2}{n^2 \pi^2}}\right)$

Informal Proof
If $\alpha $ is a root of a polynomial, then $\left({1 - \dfrac x \alpha}\right)$ is a factor.

It follows that $\sin x$ might be of the form:

If this formula is true, then $A = 1$.

This is because if $x$ is small, the LHS is approximately equal to $x$ and the RHS is approximately equal to $A x$.

This of course is not a proof.

Euler's Proof using De Moivre's Formula
Euler proved it in vol. 1 of his 1748 work Introductio in analysin infinitorum using De Moivre's Formula:


 * $\sin x = \dfrac {\left({\cos \dfrac x n + i \sin \dfrac x n}\right)^n - \left({\cos \dfrac x n - i \sin \dfrac x n}\right)^n} {2i}$

The difference between two $n$th powers can be extracted into linear factors using $n$-th roots of unity.

For large $n$, we can replace:
 * $\cos \dfrac x n$ by $1$


 * $\sin \dfrac x n$ by $\dfrac x n$

Proof without Complex Numbers
Euler's use of complex numbers can be avoided as follows.

For odd $n$, we have that $\sin x$ is a polynomial of degree $n$ in $\sin \dfrac x n$.

The roots of this polynomial are the numbers $\sin \dfrac {k \pi} n$ where $k$ is any integer.

The result follows from:


 * Factoring the polynomial
 * making $n$ go to infinity
 * replacing $\sin y$ by $y$ for small $y$.