Proof of the Viète formula

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


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} .


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).

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