Proof of the Viète formula

The Viète formula is the following infinite product type representation of the mathematical constant &pi;:


 * $$\frac2\pi=

\frac{\sqrt2}2 \frac{\sqrt{2+\sqrt2}}2 \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots$$

The expression on the right hand side has to be understood as a limit expression (as $$ n \rightarrow \infty $$)


 * $$\lim_{n \rightarrow \infty} \prod_{i=1}^n {a_i \over 2}

$$ where an is the nested quadratic radical given by the recursion $$ a_n=\sqrt{2+a_{n-1}} $$ with initial condition $$ a_1=\sqrt{2} $$.

Proof
Using an iterated application of the double-angle formula


 * $$\, \sin(2x)=2\sin(x)\cos(x) $$

for sine one first proves the identity
 * $$ {{\sin(2^n x)}\over {2^n \sin(x)}}=\prod_{i=0}^{n-1} \cos(2^i x)

$$

valid for all positive integers n. Letting x=y/2n and dividing both sides by cos(y/2) yields


 * $$ {{\sin( y)}\over {\cos({y\over 2} )}}\cdot{1\over {2^n \sin({y\over {2^n}})}}=\prod_{i=1}^{n-1} \cos\left({y\over {2^{i+1}}}\right).

$$

Using the double-angle formula sin y=2sin(y/2)cos(y/2) again gives


 * $$ {{2\sin({y\over 2})}\over {2^n \sin({y\over {2^n}})}}=\prod_{i=1}^{n-1} \cos\left({y\over {2^{i+1}}}\right).

$$

Substituting y=&pi; gives the identity


 * $$ {2\over {2^n \sin({\pi \over {2^n}})}}=\prod_{i=2}^{n} \cos\left({\pi\over {2^i}} \right) \.

$$

It remains to match the factors on the right-hand side of this identity with the terms an. Using the half-angle formula for cosine,


 * $$2\cos(x/2)=\sqrt{2+2\cos x},$$

one derives that $$ b_i=2\cos\left({\pi\over {2^{i+1}}}\right) $$ satisfies the recursion $$ \,b_{i+1}=\sqrt{2+b_i} $$ with initial condition $$ b_1= 2\cos\left({\pi \over 4}\right)=\sqrt{2}=a_1 $$. Thus an=bn for all positive integers n.

The Viète formula now follows by taking the limit n &rarr; &infin;. Note here that


 * $$ \lim_{n \rightarrow \infty} {2\over {2^n \sin({\pi \over {2^n}})}}={2\over \pi}  $$

as a consequence of the fact that $$ \lim_{x\rightarrow 0} \,{x\over {\sin x}}=1 $$ (this follows from l'Hôpital's rule).