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 $y$.

Proof by Integration
For $x \in \R$ and $n \in \N$, let:


 * $\displaystyle I_n \left({x}\right) = \int_0^{\pi / 2} \cos {x t} \cos^n t \ \mathrm d t $

Observe that $I_0 \left({0}\right) = \dfrac {\pi} 2$ and:

which yields:


 * $(1): \quad \sin \left({\dfrac {\pi x} 2}\right) = \dfrac {\pi x} 2 \dfrac {I_0 \left({x}\right)} {I_0 \left({0}\right)}$

Integrating by parts twice with $n \ge 2$, we have:

which yields the reduction formula:


 * $n \left({n - 1}\right) I_{n - 2} \left({x}\right) = \left({n^2 - x^2}\right) I_n \left({x}\right)$

Substituting $x = 0$ we obtain:


 * $n \left({n - 1}\right) I_{n - 2} \left({0}\right) = n^2 I_n \left({0}\right)$

From Shape of Cosine Function, it is clear that $I_n \left({0}\right) > 0$ for $n \ge 0 $.

Therefore we can divide the two equations to get:


 * $(2): \quad \dfrac {I_{n - 2} \left({x}\right)} {I_{n - 2} \left({0}\right)} = \left({1 - \dfrac {x^2} {n^2} }\right) \dfrac {I_n \left({x}\right)} {I_n \left({0}\right)}$

By Relative Sizes of Definite Integrals we have:

which yields the inequality:


 * $\left \vert{1 - \dfrac {I_n \left({x}\right)} {I_n \left({0}\right)} }\right \vert \le \dfrac {x^2} {2 n}$

It follows from Squeeze Theorem that:


 * $\displaystyle (3): \quad \lim_{n \to \infty} \frac {I_n \left({x}\right)} {I_n \left({0}\right)} = 1$

Consider the equation, for even $n$:


 * $\displaystyle \sin \left({\frac {\pi x} 2}\right) = \frac {\pi x} 2 \prod_{i \mathop = 1}^{n / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_n \left({x}\right)} {I_n \left({0}\right)}$

This is true for $n = 0$ by $(1)$.

Suppose it is true for some $n = k$.

Then:

So it is true for all even $n$ by induction.

Taking the limit as $n \to \infty$ we have:

or equivalently, letting $\dfrac {\pi x} 2 \mapsto x$:


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