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: {{begin-eqn}} {{eqn | l = \map {I_0} z       | r = \int_0^{\pi / 2} \cos {z t} \rd t       | c =  }} {{eqn | r = \frac 1 z \map \sin {\frac {\pi z} 2}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = z \map {I_n} z       | r = n \int_0^{\pi / 2} \sin {z t} \cos^{n - 1} t \sin t \rd t        | c =  }} {{eqn | l = z^2 \map {I_n} z       | r = n \int_0^{\pi / 2} \cos {z 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 {z 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} z - n \paren {n - 1} \map {I_{n - 2} } z       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \cmod {1 - \cos {z t} }       | r = \cmod {1 - \map \cos {x + i y} t}       | c =  }} {{eqn | r = \cmod {1 - \cos {x t} \cos {i y t} + \sin x t \sin {i y t} }       | c = Cosine of Sum }} {{eqn | r = \cmod {1 - \cos {x t} \cosh {y t} + i \sin {x t} \sinh {y t} }       | c = Hyperbolic Sine in terms of Sine and Hyperbolic Cosine in terms of Cosine }} {{eqn | r = \sqrt {\paren {1 - \cos {x t} \cosh {y t} }^2 + \paren {\sin {x t} \sinh {y t} }^2}       | c =  }} {{eqn | r = \sqrt {4 \map {\sinh^4} {\frac {y t} 2} + 8 \map {\sinh^2} {\frac {y t} 2} \map {\sin^2} {\frac {x t} 2} + 4 \map {\sin^4} {\frac {x t} 2} }       | c = Double Angle Formulas }} {{eqn | r = 2 \map {\sinh^2} {\frac {y t} 2} + 2 \sin^2 {\frac {x t} 2}       | c =  }} {{end-eqn}} 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 $t \in \closedint 0 {\dfrac {\pi} 2}$: {{begin-eqn}} {{eqn | o = \le       | l = \frac {\map \sinh {y t / 2} } {y t / 2}       | r = \frac {\map \sinh {y \pi / 4} } {y \pi /  4}       | c =  }} {{eqn | ll= \leadsto       | o = \le       | l = \frac {\map {\sinh^2} {y t / 2} } {\paren {y t / 2}^2}       | r = \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2}       | c =  }} {{eqn | ll= \leadsto       | o = \le       | l = 2 \map {\sinh^2} {y t / 2}       | r = \frac 1 2 \frac {\map {\sinh^2} {y \pi / 4} } {\paren {y \pi / 4}^2} t^2       | c = }} {{end-eqn}} So we deduce: {{begin-eqn}} {{eqn | o = \le       | l = \cmod {1 - \cos {z t} }       | r = \frac 1 2 \paren {x^2 + \frac {\map {\sinh^2} {y \pi / 4} } {\paren {\pi / 4}^2} } t^2       | c =  }} {{eqn | r = \map C {x, y} t^2       | c = where $C$ is a non-negative function of $x$ and $y$  }} {{end-eqn}} By Relative Sizes of Definite Integrals we have: {{begin-eqn}} {{eqn | l = \cmod {\map {I_n} 0 - \map {I_n} z}       | r = \cmod {\int_0^{\pi / 2} \paren {1 - \cos {z t} } \cos^n t \rd t}       | c =  }} {{eqn | o = \le       | r = \map C {x, y} \int_0^{\pi / 2} t^2 \cos^n t \rd t       | c = Cosine Inequality }} {{eqn | o = \le       | r = \map C {x, y} \int_0^{\pi / 2} t \cos^{n - 1} t \sin t \rd t       | c = Tangent Inequality }} {{eqn | r = \frac {\map C {x, y} } n \int_0^{\pi / 2} \cos^n t \rd t       | c = Integration by Parts }} {{eqn | r = \frac {\map C {x, y} } n \map {I_n} 0       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \map \sin {\frac {\pi z} 2}       | r = \frac {\pi z} 2 \prod_{i \mathop = 1}^{k / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_k} z} {\map {I_k} 0}       | c =  }} {{eqn | r = \frac {\pi z} 2 \prod_{i \mathop = 1}^{\paren {k + 2} / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_{k + 2} } z} {\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 z} 2}       | r = \lim_{n \mathop \to \infty} \frac {\pi z} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \frac {z^2} {\paren {2 i}^2} } \frac {\map {I_n} z} {\map {I_n} 0}       | c =  }} {{eqn | r = \frac {\pi z} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {z^2} {\paren {2 i}^2} } \lim_{n \mathop \to \infty} \frac {\map {I_n} z} {\map {I_n} 0}       | c = Product Rule for Limits of Functions }} {{eqn | r = \frac {\pi z} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {z^2} {\paren {2 i}^2} }       | c = by $(3)$ }} {{end-eqn}} or equivalently, letting $\dfrac {\pi z} 2 \mapsto z$: :$\ds \sin z = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2 \pi^2} }$ {{qed}}