Euler Formula for Sine Function/Complex Numbers/Proof 1

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 $n \in \N$, let:


 * $\displaystyle I_n \left({z}\right) = \int_0^{\pi / 2} \cos {z t} \cos^n t \ \mathrm d t $

Observe that $I_0 \left({0}\right) = \dfrac {\pi} 2$ and:

which yields:


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

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

which yields the reduction formula:


 * $n \left({n - 1}\right) I_{n - 2} \left({z}\right) = \left({n^2 - z^2}\right) I_n \left({z}\right)$

Substituting $z = 0$ we obtain:


 * $n \left({n - 1}\right) I_{n - 2} \left({0}\right) = n^2 I_n \left({0}\right)$

From Shape of Cosine Function, it is clear that $I_n \left({0}\right) > 0$ for $n \ge 0 $.

Therefore we can divide the two equations to get:


 * $(2): \quad \dfrac {I_{n - 2} \left({z}\right)} {I_{n - 2} \left({0}\right)} = \left({1 - \dfrac {z^2} {n^2} }\right) \dfrac {I_n \left({z}\right)} {I_n \left({0}\right)}$

We have:

From Sine Inequality we have that $2 \sin^2 \left({\dfrac {x t} 2}\right) \le \dfrac 1 2 x^2 t^2$.

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

So we deduce:

By Relative Sizes of Definite Integrals we have:

which yields the inequality:


 * $\left \vert{1 - \dfrac {I_n \left({z}\right)} {I_n \left({0}\right)} }\right \vert \le \dfrac {C \left({x,y}\right)} n$

It follows from Squeeze Theorem that:


 * $\displaystyle (3): \quad \lim_{n \to \infty} \frac {I_n \left({z}\right)} {I_n \left({0}\right)} = 1$

Consider the equation, for even $n$:


 * $\displaystyle \sin \left({\frac {\pi z} 2}\right) = \frac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \left({1 - \frac {z^2} {\left({2 i}\right)^2} }\right) \frac {I_n \left({z}\right)} {I_n \left({0}\right)}$

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

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

Then:

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

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

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


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