Euler Formula for Sine Function/Complex Numbers/Proof 2

Theorem

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

for all $z \in \C$.

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


 * $\displaystyle f_n \left({z}\right) = \frac 1 2 \left[{\left({1 + \frac z n}\right)^n - \left({1 - \frac z n}\right)^n }\right]$

Then $f_n \left({z}\right) = 0$ if and only if:

Let $n = 2 m + 1$.

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


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

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

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

Hence $C = 1$, and we obtain:


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

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

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:


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

Now let $1 < l < m$.

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

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


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

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

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


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

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


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