# Euler Formula for Sine Function/Real Numbers

## Theorem

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

for all $x \in \R$.

## Proof 1

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

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

Observe that:

$I_0 \left({0}\right) = \dfrac {\pi} 2$

and:

 $\displaystyle I_0 \left({x}\right)$ $=$ $\displaystyle \int_0^{\pi / 2} \cos {x t} \rd t$ $\displaystyle$ $=$ $\displaystyle \frac 1 x \sin \left({\frac {\pi x} 2 }\right)$

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:

 $\displaystyle x I_n \left({x}\right)$ $=$ $\displaystyle n \int_0^{\pi / 2} \sin {x t} \cos^{n - 1} t \sin t \rd t$ $\displaystyle x^2 I_n \left({x}\right)$ $=$ $\displaystyle n \int_0^{\pi / 2} \cos{x t} \left({\cos^n t - \left({n - 1}\right) \cos^{n - 2} t \sin^2 t }\right) \rd t$ $\displaystyle$ $=$ $\displaystyle n \int_0^{\pi / 2} \cos{x t} \left({n \cos^n t - \left({n - 1}\right) \cos^{n - 2} t }\right) \rd t$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle n^2 I_n \left({x}\right) - n \left({n - 1}\right) I_{n - 2} \left({x}\right)$

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:

 $\displaystyle \left \vert{I_n \left({0}\right) - I_n \left({x}\right)}\right \vert$ $=$ $\displaystyle \int_0^{\pi / 2} \left({1 - \cos {x t} }\right) \cos^n t \rd t$ the integral is non-negative $\displaystyle$ $\le$ $\displaystyle \frac {x^2} 2 \int_0^{\pi / 2} t^2 \cos^n t \rd t$ Cosine Inequality $\displaystyle$ $\le$ $\displaystyle \frac {x^2} 2 \int_0^{\pi / 2} t \cos^{n - 1} t \sin t \rd t$ Tangent Inequality $\displaystyle$ $=$ $\displaystyle \frac {x^2} {2 n} \int_0^{\pi / 2} \cos^n t \r d t$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^2} {2 n} I_n \left({0}\right)$

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:

 $\displaystyle \sin \left({\frac {\pi x} 2}\right)$ $=$ $\displaystyle \frac {\pi x} 2 \prod_{i \mathop = 1}^{k / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_k \left({x}\right)} {I_k \left({0}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {\pi x} 2 \prod_{i \mathop = 1}^{\left({k + 2}\right) / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_{k + 2} \left({x}\right)} {I_{k + 2} \left({0}\right)}$ by $(2)$

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

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

 $\displaystyle \sin \left({\frac {\pi x} 2}\right)$ $=$ $\displaystyle \lim_{n \to \infty} \frac {\pi x} 2 \prod_{i \mathop = 1}^{n / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_n \left({x}\right)} {I_n \left({0}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \lim_{n \to \infty} \frac {I_n \left({x}\right)} {I_n \left({0}\right)}$ Product Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right)$ by $(3)$

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

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

$\blacksquare$

## Proof 2

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 3

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:

$\displaystyle \sin x = A x \prod \left({1 - \frac x {k \pi} }\right)$

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

The intuition is as follows.

 $\displaystyle \sin x$ $=$ $\displaystyle \ldots \left({1 - \frac x {2 \pi} }\right) \left({1 - \frac x \pi}\right) A x \left({1 + \frac x \pi}\right) \left({1 + \frac x {2 \pi} }\right)\cdots$ $\displaystyle$ $=$ $\displaystyle A x \left({1 - \frac {x^2} {\pi^2} }\right) \left({1 - \frac {x^2} {2^2 \pi^2} }\right) \left({1 - \frac {x^2} {3^2 \pi^2} }\right) \cdots$ $\displaystyle \implies \ \$ $\displaystyle \frac {\sin x} x$ $=$ $\displaystyle A \left({1 - \frac {x^2} {\pi^2} }\right) \left({1 - \frac {x^2} {2^2 \pi^2} }\right) \left({1 - \frac {x^2} {3^2 \pi^2} }\right) \cdots$ for $x \ne 0$.

That $\dfrac {\sin x} x \to 1$ as $x \to 0$ is a well known limit.

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

We now formalize the above claims.

## Proof 4

For $x \in \R$ and $n \in \N_{> 0}$, let:

$\map {f_n} x = \dfrac 1 2 \paren {\paren {1 + \dfrac x n}^n - \paren {1 - \dfrac x n}^n }$

Then $\map {f_n} x = 0$ if and only if:

 $\displaystyle \paren {1 + \frac x n}^n$ $=$ $\displaystyle \paren {1 - \frac x n}^n$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 1 + \frac x n$ $=$ $\displaystyle \paren {1 - \frac x n} e^{2 \pi i \frac k n}$ for $k \in \Z$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x$ $=$ $\displaystyle n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1}$ $\displaystyle$ $=$ $\displaystyle n i \, \map \tan {\frac {k \pi} n }$ Tangent Exponential Formulation

Hence the roots of $\map {f_{2 m + 1} } x$ are:

$\paren {2 m + 1} i \, \map \tan {\dfrac {k \pi} {2 m + 1} }$

for $-m \le k \le m$.

Observe that $\map {f_{2 m + 1} } x$ is a polynomial of degree $2 m + 1$.

Then for some constant $C$, we have:

 $\displaystyle \map {f_{2 m + 1} } x$ $=$ $\displaystyle C x \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \paren {1 - \frac x {\paren {2 m + 1} i \, \map \tan {k \pi / \paren {2 m + 1} } } }$ Polynomial Factor Theorem $\displaystyle$ $=$ $\displaystyle C x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$ Tangent Function is Odd

It can be seen from the Binomial Theorem that the coefficient of $x$ in $\map {f_{2 m + 1} } x$ is $1$.

Hence $C = 1$, and we obtain:

$\displaystyle \map {f_{2 m + 1} } x = x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$

Let $l < m$.

Then:

$\displaystyle x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$

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

 $\displaystyle \lim_{m \mathop \to \infty} x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} \paren {\frac {k \pi / \paren {2 m + 1} } {\map \tan {k \pi / \paren {2 m + 1} } } }^2 }$ $\le$ $\displaystyle \frac 1 2 \paren {e^x - e^{-x} }$ Definition of Exponential Function $\displaystyle \leadsto \ \$ $\displaystyle x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} }$ $\le$ $\displaystyle \sinh x$ Limit of Tan X over X and Definition of Hyperbolic Sine

By Tangent Inequality, we have:

$\map \tan {\dfrac {k \pi} {2 m + 1} } \le \dfrac {k \pi} {2 m + 1}$

and hence:

$\displaystyle \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$

Taking the limit as $l \to \infty$, we have by the Squeeze Theorem:

$\displaystyle x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$

Substituting $x \mapsto i x$, we obtain:

$\displaystyle \sin x = x \prod_{k \mathop = 1}^\infty \paren {1 - \frac {x^2} {k^2 \pi^2} }$

$\blacksquare$