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: {{begin-eqn}} {{eqn | l = I_0 \left({x}\right)       | r = \int_0^{\pi / 2} \cos {x t} \rd t       | c =  }} {{eqn | r = \frac 1 x \sin \left({\frac {\pi x} 2 }\right)       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = x I_n \left({x}\right)       | r = n \int_0^{\pi / 2} \sin {x t} \cos^{n - 1} t \sin t \rd t        | c =  }} {{eqn | l = x^2 I_n \left({x}\right)       | r = n \int_0^{\pi / 2} \cos{x t} \left({\cos^n t - \left({n - 1}\right) \cos^{n - 2} t \sin^2 t }\right) \rd t       | c =  }} {{eqn | r = n \int_0^{\pi / 2} \cos{x t} \left({n \cos^n t - \left({n - 1}\right) \cos^{n - 2} t }\right) \rd t       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = n^2 I_n \left({x}\right) - n \left({n - 1}\right) I_{n - 2} \left({x}\right)       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \left \vert{I_n \left({0}\right) - I_n \left({x}\right)}\right \vert       | r = \int_0^{\pi / 2} \left({1 - \cos {x t} }\right) \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 }} {{eqn | o = \le       | r = \frac {x^2} 2 \int_0^{\pi / 2} t \cos^{n - 1} t \sin t \rd t       | c = Tangent Inequality }} {{eqn | r = \frac {x^2} {2 n} \int_0^{\pi / 2} \cos^n t \r d t       | c = Integration by Parts }} {{eqn | r = \frac {x^2} {2 n} I_n \left({0}\right)       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \sin \left({\frac {\pi x} 2}\right)       | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^{k / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_k \left({x}\right)} {I_k \left({0}\right)}       | c =  }} {{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^{\left({k + 2}\right) / 2} \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \frac {I_{k + 2} \left({x}\right)} {I_{k + 2} \left({0}\right)}       | 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 = \sin \left({\frac {\pi x} 2}\right)       | r = \lim_{n \to \infty} \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)}       | c =  }} {{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right) \lim_{n \to \infty} \frac {I_n \left({x}\right)} {I_n \left({0}\right)}       | c = Product Rule for Limits of Functions }} {{eqn | r = \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \left({1 - \frac {x^2} {\left({2 i}\right)^2} }\right)       | c = by $(3)$ }} {{end-eqn}} or equivalently, letting $\dfrac {\pi x} 2 \mapsto x$: :$\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \left({1 - \frac {x^2} {n^2 \pi^2} }\right)$ {{qed}}