Euler Formula for Sine Function/Complex Numbers/Proof 2

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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

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:

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

$\blacksquare$