Euler Formula for Sine Function/Real Numbers

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Theorem

\(\ds \sin x\) \(=\) \(\ds x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }\)
\(\ds \) \(=\) \(\ds x \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \dotsm\)

for all $x \in \R$.


Proof 1

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

$\ds \map {I_n} x = \int_0^{\pi / 2} \cos {x t} \cos^n t \rd t $

Observe that:

$\map {I_0} 0 = \dfrac {\pi} 2$

and:

\(\ds \map {I_0} x\) \(=\) \(\ds \int_0^{\pi / 2} \cos {x t} \rd t\)
\(\ds \) \(=\) \(\ds \frac 1 x \map \sin {\frac {\pi x} 2}\)

which yields:

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


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

\(\ds x \map {I_n} x\) \(=\) \(\ds n \int_0^{\pi / 2} \sin {x t} \cos^{n - 1} t \sin t \rd t\)
\(\ds x^2 \map {I_n} x\) \(=\) \(\ds n \int_0^{\pi / 2} \cos{x t} \paren {\cos^n t - \paren {n - 1} \cos^{n - 2} t \sin^2 t} \rd t\)
\(\ds \) \(=\) \(\ds n \int_0^{\pi / 2} \cos{x t} \paren {n \cos^n t - \paren {n - 1} \cos^{n - 2} t} \rd t\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds n^2 \map {I_n} x - n \paren {n - 1} \map {I_{n - 2} } x\)

which yields the reduction formula:

$n \paren {n - 1} \map {I_{n - 2} } x = \paren {n^2 - x^2} \map {I_n} x$

Substituting $x = 0$ we obtain:

$n \paren {n - 1} \map {I_{n - 2} } 0 = n^2 \map {I_n} 0$

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

Therefore we can divide the two equations to get:

$(2): \quad \dfrac {\map {I_{n - 2} } x} {\map {I_{n - 2} } 0} = \paren {1 - \dfrac {x^2} {n^2} } \dfrac {\map {I_n} x} {\map {I_n} 0}$


By Relative Sizes of Definite Integrals we have:

\(\ds \size {\map {I_n} 0 - \map {I_n} x}\) \(=\) \(\ds \int_0^{\pi / 2} \paren {1 - \cos {x t} } \cos^n t \rd t\) the integral is non-negative
\(\ds \) \(\le\) \(\ds \frac {x^2} 2 \int_0^{\pi / 2} t^2 \cos^n t \rd t\) Cosine Inequality: $1 - \cos x t \le \dfrac {x^2 t^2} 2$
\(\ds \) \(\le\) \(\ds \frac {x^2} 2 \int_0^{\pi / 2} t \tan t \cos^n t \rd t\) Tangent Inequality: $t \le \tan t$
\(\ds \) \(=\) \(\ds \frac {x^2} 2 \int_0^{\pi / 2} t \sin t \cos^{n - 1} t \rd t\)
\(\ds \) \(=\) \(\ds \frac {x^2} {2 n} \int_0^{\pi / 2} \cos^n t \rd t\) Integration by Parts: $u = t$ and $\rd v = \sin t \cos^{n - 1} t \rd t$
\(\ds \) \(=\) \(\ds \frac {x^2} {2 n} \map {I_n} 0\)

which yields the inequality:

$\size {1 - \dfrac {\map {I_n} x} {\map {I_n} 0} } \le \dfrac {x^2} {2 n}$

It follows from Squeeze Theorem that:

$\ds (3): \quad \lim_{n \mathop \to \infty} \frac {\map {I_n} x} {\map {I_n} 0} = 1$


Consider the equation, for even $n$:

$\ds \map \sin {\frac {\pi x} 2} = \frac {\pi x} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_n} x} {\map {I_n} 0}$

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

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

Then:

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

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

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

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

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

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

$\blacksquare$


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 $\dfrac x n$


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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:

$\ds \sin x = A x \prod \paren {1 - \frac x {k \pi} }$

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

The intuition is as follows.

\(\ds \sin x\) \(=\) \(\ds \ldots \paren {1 - \frac x {2 \pi} } \paren {1 - \frac x \pi} A x \paren {1 + \frac x \pi} \paren {1 + \frac x {2 \pi} } \cdots\)
\(\ds \) \(=\) \(\ds A x \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots\)
\(\ds \leadsto \ \ \) \(\ds \frac {\sin x} x\) \(=\) \(\ds A \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots\) for $x \ne 0$

From Limit of $\dfrac {\sin x} x$ at Zero:

$\dfrac {\sin x} x \to 1$ as $x \to 0$

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

We now formalize the above claims.


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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:

\(\ds \paren {1 + \frac x n}^n\) \(=\) \(\ds \paren {1 - \frac x n}^n\)
\(\ds \leadstoandfrom \ \ \) \(\ds 1 + \frac x n\) \(=\) \(\ds \paren {1 - \frac x n} e^{2 \pi i \frac k n}\) for $k \in \Z$
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(=\) \(\ds n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1}\)
\(\ds \) \(=\) \(\ds n i \, \map \tan {\frac {k \pi} n }\) Euler's Tangent Identity


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:

\(\ds \map {f_{2 m + 1} } x\) \(=\) \(\ds C x \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \paren {1 - \frac x {\paren {2 m + 1} i \, \map \tan {k \pi / \paren {2 m + 1} } } }\) Polynomial Factor Theorem
\(\ds \) \(=\) \(\ds C x \prod_{k \mathop = 1}^m \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }\) 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:

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


Let $l < m$.

Then:

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

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

\(\ds \lim_{m \mathop \to \infty} x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} \paren {\frac {k \pi / \paren {2 m + 1} } {\map \tan {k \pi / \paren {2 m + 1} } } }^2 }\) \(\le\) \(\ds \frac 1 2 \paren {e^x - e^{-x} }\) Definition of Exponential Function
\(\ds \leadsto \ \ \) \(\ds x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} }\) \(\le\) \(\ds \sinh x\) Limit of $\dfrac {\tan x} x$ at Zero and Definition of Hyperbolic Sine


By Tangent Inequality, we have:

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

and hence:

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


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

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

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

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

$\blacksquare$

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