Theorem: Euler Formula for Sine Function/Real Numbers

{{begin-eqn}}
{{eqn | l = \sin x
      | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }
      | c = 
}}
{{eqn | r = x \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \dotsm
      | c = 
}}
{{end-eqn}}
for all $x \in \R$.

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:
{{begin-eqn}}
{{eqn | l = \map {I_0} x
      | r = \int_0^{\pi / 2} \cos {x t} \rd t
      | c = 
}}
{{eqn | r = \frac 1 x \map \sin {\frac {\pi x} 2}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = x \map {I_n} x
      | r = n \int_0^{\pi / 2} \sin {x t} \cos^{n - 1} t \sin t \rd t 
      | c = 
}}
{{eqn | l = x^2 \map {I_n} x
      | r = n \int_0^{\pi / 2} \cos{x t} \paren {\cos^n t - \paren {n - 1} \cos^{n - 2} t \sin^2 t} \rd t
      | c = 
}}
{{eqn | r = n \int_0^{\pi / 2} \cos{x t} \paren {n \cos^n t - \paren {n - 1} \cos^{n - 2} t} \rd t
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = n^2 \map {I_n} x - n \paren {n - 1} \map {I_{n - 2} } x
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \size {\map {I_n} 0 - \map {I_n} x}
      | r = \int_0^{\pi / 2} \paren {1 - \cos {x t} } \cos^n t \rd t
      | c = the integral is non-negative
}}
{{eqn | o = \le
      | r = \frac {x^2} 2 \int_0^{\pi / 2} t^2 \cos^n t \rd t
      | c = Cosine Inequality: $1 - \cos x t \le \dfrac {x^2 t^2} 2$
}}
{{eqn | o = \le
      | r = \frac {x^2} 2 \int_0^{\pi / 2} t \tan t \cos^n t \rd t
      | c = Tangent Inequality: $t \le \tan t$
}}
{{eqn | r = \frac {x^2} 2 \int_0^{\pi / 2} t \sin t \cos^{n - 1} t \rd t
      | c = 
}}
{{eqn | r = \frac {x^2} {2 n} \int_0^{\pi / 2} \cos^n t \rd t
      | c = Integration by Parts: $u = t$ and $\rd v = \sin t \cos^{n - 1} t \rd t$
}}
{{eqn | r = \frac {x^2} {2 n} \map {I_n} 0
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map \sin {\frac {\pi x} 2}
      | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^{k / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_k} x} {\map {I_k} 0}
      | c = 
}}
{{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^{\paren {k + 2} / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_{k + 2} } x} {\map {I_{k + 2} } 0}
      | c = by $(2)$
}}
{{end-eqn}}
So it is true for all even $n$ by induction.
Taking the limit as $n \to \infty$ we have:
{{begin-eqn}}
{{eqn | l = \map \sin {\frac {\pi x} 2}
      | r = \lim_{n \mathop \to \infty} \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}
      | c = 
}}
{{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {x^2} {\paren {2 i}^2} } \lim_{n \mathop \to \infty} \frac {\map {I_n} x} {\map {I_n} 0}
      | c = Product Rule for Limits of Real Functions
}}
{{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {x^2} {\paren {2 i}^2} }
      | c = by $(3)$
}}
{{end-eqn}}
or equivalently, letting $\dfrac {\pi x} 2 \mapsto x$:
:$\ds \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$
{{qed}}