Euler Formula for Sine Function/Complex Numbers/Proof 1

Proof
For $z \in \C$ and $n \in \N$, let:


 * $\ds \map {I_n} z = \int_0^{\pi / 2} \cos {z t} \cos^n t \rd t $

Observe that $\map {I_0} 0 = \dfrac {\pi} 2$ and:

which yields:


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

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

which yields the reduction formula:


 * $n \paren {n - 1} \map {I_{n - 2} } z = \paren {n^2 - z^2} \map {I_n} z$

Substituting $z = 0$ we obtain:


 * $n \paren {n - 1} \map {I_{n - 2} } 0 = n^2 \map {I_n} 0$

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

Therefore we can divide the two equations to get:


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

We have:

From Sine Inequality we have that $2 \map {\sin^2} {\dfrac {x t} 2} \le \dfrac 1 2 x^2 t^2$.

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

So we deduce:

By Relative Sizes of Definite Integrals we have:

which yields the inequality:


 * $\cmod {1 - \dfrac {\map {I_n} z} {\map {I_n} 0} } \le \dfrac {\map C {x, y} } n$

It follows from Squeeze Theorem that:


 * $(3): \quad \ds \lim_{n \mathop \to \infty} \frac {\map {I_n} z} {\map {I_n} 0} = 1$

Consider the equation, for even $n$:


 * $\ds \map \sin {\dfrac {\pi z} 2} = \dfrac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \dfrac {z^2} {\paren {2 i}^2} } \dfrac {\map {I_n} z} {\map {I_n} 0}$

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 $\dfrac {\pi z} 2 \mapsto z$:


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