Trigonometry/Price's Theorem

Price's Theorem states that as $n\to \mathrm{\infty }$

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

Lemma

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

Proof of lemma

As $n\to \mathrm{\infty },\frac{\theta }{2^n}\to 0$ hence $\frac{\sin \left(\frac{\theta }{2^n}\right)}{\frac{\theta }{2^n}}\to 1$. 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).