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).