# Trigonometry/Price's Theorem

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

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

Lemma

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

Proof of lemma

As $n \rightarrow \inf, \, \frac{\theta}{2^n} \rightarrow 0$ 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).