Euler Formula for Sine Function/Complex Numbers

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Theorem

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

for all $z \in \C$.


Proof 1

For $z \in \C$ and $n \in \N$, let:

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

Observe that $\map {I_0} 0 = \dfrac {\pi} 2$ and:

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

which yields:

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


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

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

which yields the reduction formula:

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

Substituting $z = 0$ 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:

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


We have:

\(\ds \cmod {1 - \cos {z t} }\) \(=\) \(\ds \cmod {1 - \map \cos {x + i y} t}\)
\(\ds \) \(=\) \(\ds \cmod {1 - \cos {x t} \cos {i y t} + \sin x t \sin {i y t} }\) Cosine of Sum
\(\ds \) \(=\) \(\ds \cmod {1 - \cos {x t} \cosh {y t} + i \sin {x t} \sinh {y t} }\) Hyperbolic Sine in terms of Sine and Hyperbolic Cosine in terms of Cosine
\(\ds \) \(=\) \(\ds \sqrt {\paren {1 - \cos {x t} \cosh {y t} }^2 + \paren {\sin {x t} \sinh {y t} }^2}\)
\(\ds \) \(=\) \(\ds \sqrt {4 \map {\sinh^4} {\frac {y t} 2} + 8 \map {\sinh^2} {\frac {y t} 2} \map {\sin^2} {\frac {x t} 2} + 4 \map {\sin^4} {\frac {x t} 2} }\) Double Angle Formulas
\(\ds \) \(=\) \(\ds 2 \map {\sinh^2} {\frac {y t} 2} + 2 \sin^2 {\frac {x t} 2}\)

From Sine Inequality we have that $2 \map {\sin^2} {\dfrac {x t} 2} \le \dfrac 1 2 x^2 t^2$.

By Lemma 1, $\dfrac {\sinh x} x$ is an increasing function for $x \ge 0$, so for $t \in \closedint 0 {\dfrac {\pi} 2}$:

\(\ds \frac {\map \sinh {y t / 2} } {y t / 2}\) \(\le\) \(\ds \frac {\map \sinh {y \pi / 4} } {y \pi / 4}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\map {\sinh^2} {y t / 2} } {\paren {y t / 2}^2}\) \(\le\) \(\ds \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2}\)
\(\ds \leadsto \ \ \) \(\ds 2 \map {\sinh^2} {y t / 2}\) \(\le\) \(\ds \frac 1 2 \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2} t^2\)

So we deduce:

\(\ds \cmod {1 - \cos {z t} }\) \(\le\) \(\ds \frac 1 2 \paren {x^2 + \frac {\map {\sinh^2} {y \pi / 4} } {\paren {\pi / 4}^2} } t^2\)
\(\ds \) \(=\) \(\ds \map C {x, y} t^2\) where $C$ is a non-negative function of $x$ and $y$

By Relative Sizes of Definite Integrals we have:

\(\ds \cmod {\map {I_n} 0 - \map {I_n} z}\) \(=\) \(\ds \cmod {\int_0^{\pi / 2} \paren {1 - \cos {z t} } \cos^n t \rd t}\)
\(\ds \) \(\le\) \(\ds \map C {x, y} \int_0^{\pi / 2} t^2 \cos^n t \rd t\) Cosine Inequality
\(\ds \) \(\le\) \(\ds \map C {x, y} \int_0^{\pi / 2} t \cos^{n - 1} t \sin t \rd t\) Tangent Inequality
\(\ds \) \(=\) \(\ds \frac {\map C {x, y} } n \int_0^{\pi / 2} \cos^n t \rd t\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {\map C {x, y} } n \map {I_n} 0\)

which yields the inequality:

$\cmod {1 - \dfrac {\map {I_n} z} {\map {I_n} 0} } \le \dfrac {\map C {x, y} } n$

It follows from Squeeze Theorem that:

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


Consider the equation, for even $n$:

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

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

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

Then:

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

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

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

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

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

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

$\blacksquare$


Proof 2

For $z \in \C$ and $n \in \N_{> 0}$, let:

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

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

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

Let $n = 2 m + 1$.

Then the roots of $\map {f_{2 m + 1} } z$ are:

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

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


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

Then for some constant $C$, we have:

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

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

Hence $C = 1$, and we obtain:

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


First we consider $z = x$ where $x$ is a non-negative real number.

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:

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

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 Squeeze Theorem:

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


Now let $1 < l < m$.

By Complex Modulus of Product of Complex Numbers and the Triangle Inequality, we can deduce:

\(\ds \) \(\) \(\ds \cmod {\map {f_{2 m + 1} } z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } }\)
\(\ds \) \(=\) \(\ds \cmod z \cmod {\prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } } \cdot \cmod {\prod_{k \mathop = l + 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } - 1}\)
\(\ds \) \(\le\) \(\ds \cmod z \paren {\prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } } \cdot \paren {\prod_{k \mathop = l + 1}^m \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } - 1}\)
\(\ds \) \(=\) \(\ds \map {f_{2 m + 1} } {\cmod z} - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }\)

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

$\ds \cmod {\sinh z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {k^2 \pi^2} } } \le \sinh \cmod z - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {k^2 \pi^2} }$


Now take the limit as $l \to \infty$.

By $(1)$ and Squeeze Theorem, we have:

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

Finally, substituting $z \mapsto i z$, we obtain:

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

$\blacksquare$

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