Euler Formula for Sine Function/Real Numbers/Proof 1

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


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

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

and:

which yields:


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

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

which yields the reduction formula:


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

Substituting $x = 0$ 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:


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

By Relative Sizes of Definite Integrals we have:

which yields the inequality:


 * $\left \vert{1 - \dfrac {I_n \left({x}\right)} {I_n \left({0}\right)} }\right \vert \le \dfrac {x^2} {2 n}$

It follows from Squeeze Theorem that:


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

Consider the equation, for even $n$:


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

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


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