# Vieta's Formula for Pi

(Redirected from Viète's Formula for Pi)

This proof is about an approximation formula for pi. For other uses, see Viète's Formulas.

## Theorem

$\pi = 2 \times \dfrac 2 {\sqrt 2} \times \dfrac 2 {\sqrt {2 + \sqrt 2} } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt 2} } } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } } \times \cdots$

## Proof

 $\displaystyle 1$ $=$ $\displaystyle \sin \frac \pi 2$ Sine of Half-Integer Multiple of Pi $\displaystyle$ $=$ $\displaystyle 2 \sin \frac \pi 4 \cos \frac \pi 4$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle 2 \paren {2 \sin \frac \pi 8 \cos \frac \pi 8} \cos \frac \pi 4$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle 2 \paren {2 \paren {2 \sin \frac \pi {16} \cos \frac \pi {16} } \cos \frac \pi 8} \cos \frac \pi 4$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle \cdots$ $\displaystyle$ $=$ $\displaystyle 2^{n - 1} \sin \frac \pi {2^n} \cos \frac \pi {2^n} \cos \frac \pi {2^{n - 1} } \cdots \cos \frac \pi 8 \cos \frac \pi 4$

Thus:

 $\displaystyle \frac 1 {2^{n - 1} \sin \frac \pi {2^n} }$ $=$ $\displaystyle \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac 2 \pi \times \frac {\pi / 2^n} {\map \sin {\pi / 2^n} }$ $=$ $\displaystyle \cos \frac \pi 4 \cos \frac \pi 8 \cos \frac \pi {16} \cdots \cos \frac \pi {2^{n - 1} } \cos \frac \pi {2^n}$

Then we have from the Half Angle Formula for Cosine that:

 $\displaystyle \cos \frac \pi {2^{k} }$ $=$ $\displaystyle \frac {\sqrt {2 + 2 \map \cos {\pi / 2^{k - 1} } } } 2$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 2} } } } } 2$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {2 + \sqrt {2 + \sqrt {2 + 2 \map \cos {\pi / 2^{k - 3} } } } } } 2$

So we can replace all the instances of $\cos \dfrac \pi 4$, $\cos \dfrac \pi 8$, etc. with their expansions in square roots of $2$.

Finally, we note that from Limit of Sine of X over X we have:

$\displaystyle \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$

As $n \to \infty$, then, we have that $\dfrac \pi {2^n} \to 0$, and so:

$\displaystyle \lim_{n \mathop \to \infty} \frac {\map \sin {\pi / 2^n} } {\pi / 2^n} = 1$

The result follows after some algebra.

$\blacksquare$

## Also known as

This result is also known as Vieta's (or Viète's) product.

## Source of Name

This entry was named for Franciscus Vieta.

## Historical Note

Franciscus Vieta, or François Viète as he is otherwise known, published his formula for $\pi$ in $1592$.