Euler Formula for Sine Function

Theorem

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

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:

	\(\displaystyle \)	\(\displaystyle \sin x\)	\(=\)	\(\displaystyle A x \left({1 - \frac x \pi}\right) \left({1 + \frac x \pi}\right) \left({1 - \frac x {2 \pi} }\right) \left({1 + \frac x {2 \pi} }\right) \cdots\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle A x \left({1 - \frac {x^2} {\pi^2} }\right) \left({1 - \frac {x^2} {4 \pi^2} }\right) \left({1 - \frac {x^2} {9 \pi^2} }\right) \cdots\)	\(\displaystyle \)

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 $y$.

Source of Name

This entry was named for Leonhard Paul Euler.