Euler Formula for Sine Function/Real Numbers/Proof 4

Proof
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 $\map {f_n} x = 0$ :

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:

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:

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:


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