Euler Formula for Sine Function/Real Numbers/Proof 1

Proof
For $x \in \R$ and $n \in \N$, let:


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

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

and:

which yields:


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

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

which yields the reduction formula:


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

Substituting $x = 0$ 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:


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

By Relative Sizes of Definite Integrals we have:

which yields the inequality:


 * $\size {1 - \dfrac {\map {I_n} x} {\map {I_n} 0} } \le \dfrac {x^2} {2 n}$

It follows from Squeeze Theorem that:


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

Consider the equation, for even $n$:


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

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


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