Price's Theorem states that as n → ∞ {\displaystyle n\rightarrow \infty }
Lemma
As n → ∞ , 2 n sin ( θ 2 n ) → θ {\displaystyle n\rightarrow \infty ,\,2^{n}\sin \left({\frac {\theta }{2^{n}}}\right)\rightarrow \theta } .
Proof of lemma
As n → ∞ , θ 2 n → 0 {\displaystyle n\rightarrow \infty ,\,{\frac {\theta }{2^{n}}}\rightarrow 0} hence sin ( θ 2 n ) θ 2 n → 1 {\displaystyle {\frac {\sin \left({\frac {\theta }{2^{n}}}\right)}{\frac {\theta }{2^{n}}}}\rightarrow 1} . Rearranging, the result follows.
Proof of theorem
sin ( θ ) = 2 sin ( θ 2 ) cos ( θ 2 ) = 2 2 sin ( θ 4 ) cos ( θ 4 ) cos ( θ 2 ) = . . . {\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
cos ( θ 2 ) cos ( θ 4 ) . . . cos ( θ 2 n ) = sin ( θ ) 2 n sin ( θ 2 n ) {\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).
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hence 
. Rearranging, the result follows.
