Euler Formula for Sine Function/Real Numbers/Proof 4

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


 * $f_n \left({x}\right) = \dfrac 1 2 \left[{\left({1 + \dfrac x n}\right)^n - \left({1 - \dfrac x n}\right)^n }\right]$

Then $f_n \left({x}\right) = 0$ :

Let $n = 2 m + 1$.

Then the roots of $f_{2 m + 1} \left({x}\right)$ are:


 * $\left({2 m + 1}\right) i \tan \left({\dfrac {k \pi} {2 m + 1}}\right)$

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

Observe that $f_{2m + 1} \left({x}\right)$ 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 $f_{2 m + 1} \left({x}\right)$ is $1$.

Hence $C = 1$, and we obtain:


 * $\displaystyle f_{2 m + 1} \left({x}\right) = x \prod_{k \mathop = 1}^m \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)$

Let $l < m$.

Then:


 * $\displaystyle x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \le f_{2 m + 1} \left({x}\right)$

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

By Tangent Inequality, we have:


 * $\tan \left({\dfrac {k \pi} {2 m + 1}}\right) \le \dfrac {k \pi} {2 m + 1}$

hence:


 * $\displaystyle f_{2 l + 1} \left({x}\right) \le x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) \le \sinh x$

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


 * $\displaystyle x \prod_{k \mathop = 1}^\infty \left({1 + \frac {x^2} {k^2 \pi^2} }\right) = \sinh x$

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


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