Trigonometry/Price's Theorem

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

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

Lemma

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

Proof of lemma

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

Proof of theorem

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

Thus

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

The result then follows from the lemma.

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