# Euler Formula for Sine Function/Real Numbers/Proof 1

## Theorem

 $\ds \sin x$ $=$ $\ds x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$ $\ds$ $=$ $\ds x \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \dotsm$

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:

 $\ds \map {I_0} x$ $=$ $\ds \int_0^{\pi / 2} \cos {x t} \rd t$ $\ds$ $=$ $\ds \frac 1 x \map \sin {\frac {\pi x} 2}$

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:

 $\ds x \map {I_n} x$ $=$ $\ds n \int_0^{\pi / 2} \sin {x t} \cos^{n - 1} t \sin t \rd t$ $\ds x^2 \map {I_n} x$ $=$ $\ds n \int_0^{\pi / 2} \cos{x t} \paren {\cos^n t - \paren {n - 1} \cos^{n - 2} t \sin^2 t} \rd t$ $\ds$ $=$ $\ds n \int_0^{\pi / 2} \cos{x t} \paren {n \cos^n t - \paren {n - 1} \cos^{n - 2} t} \rd t$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds n^2 \map {I_n} x - n \paren {n - 1} \map {I_{n - 2} } x$

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:

 $\ds \size {\map {I_n} 0 - \map {I_n} x}$ $=$ $\ds \int_0^{\pi / 2} \paren {1 - \cos {x t} } \cos^n t \rd t$ the integral is non-negative $\ds$ $\le$ $\ds \frac {x^2} 2 \int_0^{\pi / 2} t^2 \cos^n t \rd t$ Cosine Inequality: $1 - \cos x t \le \dfrac {x^2 t^2} 2$ $\ds$ $\le$ $\ds \frac {x^2} 2 \int_0^{\pi / 2} t \tan t \cos^n t \rd t$ Tangent Inequality: $t \le \tan t$ $\ds$ $=$ $\ds \frac {x^2} 2 \int_0^{\pi / 2} t \sin t \cos^{n - 1} t \rd t$ $\ds$ $=$ $\ds \frac {x^2} {2 n} \int_0^{\pi / 2} \cos^n t \rd t$ Integration by Parts: $u = t$ and $\rd v = \sin t \cos^{n - 1} t \rd t$ $\ds$ $=$ $\ds \frac {x^2} {2 n} \map {I_n} 0$

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:

 $\ds \map \sin {\frac {\pi x} 2}$ $=$ $\ds \frac {\pi x} 2 \prod_{i \mathop = 1}^{k / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_k} x} {\map {I_k} 0}$ $\ds$ $=$ $\ds \frac {\pi x} 2 \prod_{i \mathop = 1}^{\paren {k + 2} / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_{k + 2} } x} {\map {I_{k + 2} } 0}$ by $(2)$

So it is true for all even $n$ by induction.

Taking the limit as $n \to \infty$ we have:

 $\ds \map \sin {\frac {\pi x} 2}$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {\pi x} 2 \prod_{i \mathop = 1}^{n / 2} \paren {1 - \frac {x^2} {\paren {2 i}^2} } \frac {\map {I_n} x} {\map {I_n} 0}$ $\ds$ $=$ $\ds \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {x^2} {\paren {2 i}^2} } \lim_{n \mathop \to \infty} \frac {\map {I_n} x} {\map {I_n} 0}$ Product Rule for Limits of Real Functions $\ds$ $=$ $\ds \frac {\pi x} 2 \prod_{i \mathop = 1}^\infty \paren {1 - \frac {x^2} {\paren {2 i}^2} }$ by $(3)$

or equivalently, letting $\dfrac {\pi x} 2 \mapsto x$:

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

$\blacksquare$