Sin icin aci formulunu tekrar ve tekrar kullanirsak
$\sin x=2\sin(\frac{x}{2})\cos(\frac{x}{2})$
$=2.2\sin(\frac{x}{4})\cos(\frac{x}{4})\cos(\frac{x}{2})$
$=2.2.2\sin(\frac{x}{8})\cos(\frac{x}{8})\cos(\frac{x}{4})\cos(\frac{x}{2})$
$=\vdots$
$=2^n\sin(\frac{x}{2^n}) \bigg(\prod_{k=1}^n\cos(\frac{x}{2^k})\bigg)$
elde ederiz.
$lim_{n\rightarrow\infty} 2^n\sin(\frac{x}{2^n})=x $ oldugundan
$\frac{\sin x}{x}= \prod_{k=1}^{\infty}\cos(\frac{x}{2^k})$ olur..
Surdan alindi..
http://www.calpoly.edu/~kmorriso/Research/cosine.pdf