Euler Formula for Sine Function

Theorem

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

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


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

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:
 * $$\cos \frac x n$$ by $$1$$;


 * $$\sin \frac x n$$ by $$\frac x n$$.

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 $$\sin \frac x n$$.

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

The result follows from:


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