# Trigonometry/Price's Theorem

Price's Theorem states that as ${\displaystyle n\rightarrow \infty }$

${\displaystyle \cos \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{4}}\right)\cos \left({\frac {\theta }{8}}\right)...\cos \left({\frac {\theta }{2^{n}}}\right)\rightarrow {\frac {\sin \theta }{\theta }}}$.

Lemma

As ${\displaystyle n\rightarrow \infty ,\,2^{n}\sin \left({\frac {\theta }{2^{n}}}\right)\rightarrow \theta }$.

Proof of lemma

As ${\displaystyle n\rightarrow \infty ,\,{\frac {\theta }{2^{n}}}\rightarrow 0}$ hence ${\displaystyle {\frac {\sin \left({\frac {\theta }{2^{n}}}\right)}{\frac {\theta }{2^{n}}}}\rightarrow 1}$. Rearranging, the result follows.

Proof of theorem

${\displaystyle \sin(\theta )=2\sin \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{2}}\right)=2^{2}\sin \left({\frac {\theta }{4}}\right)\cos \left({\frac {\theta }{4}}\right)\cos \left({\frac {\theta }{2}}\right)=...}$

Thus

${\displaystyle \cos \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{4}}\right)...\cos \left({\frac {\theta }{2^{n}}}\right)={\frac {\sin(\theta )}{2^{n}\sin \left({\frac {\theta }{2^{n}}}\right)}}}$

The result then follows from the lemma.

This theorem is due to Bartholomew Price (1818-1898).