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The Viète formula is the following infinite product type representation of the mathematical constant π:

$\frac2\pi= \frac{\sqrt2}2 \frac{\sqrt{2+\sqrt2}}2 \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots$

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=π 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 → ∞. 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).