Viète’s Formula

May 3rd, 2017

Today we are going to derive one of the most elegant formulas math has ever seen (in my opinion), Viète’s formula. The formula is based on the following idea

$$\frac{2}{\pi} = \frac{\text{area of inscribed square}}{\text{area of unit circle}}$$

Viète realized this formula could be manipulated by adding intermediate steps between the circle and the square

$$\frac{2}{\pi} = \frac{\text{area of inscribed square}}{\text{area of inscribed octagon}}\cdot\frac{\text{area of inscribed octagon}}{\text{area of unit circle}}$$

$$\frac{2}{\pi} = \frac{\text{area of inscribed square}}{\text{area of inscribed octagon}}\cdot\frac{\text{area of inscribed octagon}}{\text{area of inscribed }2^4\text{-gon}}\cdot\frac{\text{area of inscribed }2^4\text{-gon}}{\text{area of unit circle}}$$

As we look at polygons with more and more sides, their area approaches that of the circle. In math terms, we can write

$$\lim_{n\rightarrow\infty}\frac{\text{area of inscribed }2^n\text{-gon}}{\text{area of unit circle}} = 1$$



First, we need to create some notation to help our calculations. Define $Q_n$ as

$$Q_n = \frac{\text{area of } 2^n\text{-gon}}{\text{area of } 2^{n+1}\text{-gon}}$$

We know that as $n\rightarrow\infty$, $Q_n\rightarrow 1$. As a result, the infinite product has a chance of converging (and in this case, it does!). We can then write

$$\frac{2}{\pi} = Q_{2}\cdot Q_{3} \cdot Q_{4}\dots$$


Area Computation Part 1

Consider the regular n-gon. We will compute the area by dividing the polygon into $n$ triangles.


The interior angle for each of these isosceles triangles is


We also know that the side lengths are 1, since the shape is inscribed inside a circle of radius 1. We will use some trig to get the base and height of the triangle. Here is a picture for an octagon.


$$B = 2\sin\frac{\theta}{2}\hspace{10mm}H=\cos\frac{\theta}{2}$$

$$\text{area} = \frac{BH}{2} = \sin\frac{\theta}{2}\cos\frac{\theta}{2}$$

We have $n$ triangles in our shape, and we already said that the angle $\theta=\frac{2\pi}{n}$, so

$$\text{area of inscribed n-gon} =  n\cdot\sin\frac{\pi}{n}\cos\frac{\pi}{n}$$


Area Computation Part 2

We now go back to our definition for $Q_n$

$$Q_n = \frac{\text{area of } 2^n\text{-gon}}{\text{area of } 2^{n+1}\text{-gon}}$$

$$Q_n = \frac{ 2^n\cdot\sin\frac{\pi}{2^n}\cos\frac{\pi}{2^n}}{2^{n+1}\cdot\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}$$

$$= \frac{\sin\frac{\pi}{2^n}\cos\frac{\pi}{2^n}}{2\cdot\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}$$

Now we need to make this formula look a little nicer. We can do this by using the trivial fact that

$$\frac{\pi}{2^n} = 2\cdot\frac{\pi}{2^{n+1}}$$

$$Q_n = \frac{\sin\frac{2\pi}{2^{n+1}}\cos\frac{\pi}{2^{n}}}{2\cdot\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}$$

Now use the double angle formula for $\sin$

$$\sin2\theta = 2\sin\theta\cos\theta$$

$$\sin\frac{2\pi}{2^{n+1}} = 2\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}$$

$$Q_n = \frac{2\cdot\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n}}}{2\cdot\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}$$

The ugly stuff cancels out wonderfully, leaving us with

$$ Q_n=  \cos\frac{\pi}{2^{n}}$$


$$ \bbox[15px,border:1px solid black]{\frac{2}{\pi} = \cos\frac{\pi}{2} \cos\frac{\pi}{4} \cos\frac{\pi}{8}\dots}$$



The issue with the formula above is that, well, it uses $\pi$ to compute $\pi$. This is criminal circular logic. However, we can easily remedy this. Let’s look at the double angle formula for $\cos$

$$\cos 2\theta = 2\cos^2\theta – 1$$

$$\cos\theta = 2\cos^2\frac{\theta}{2} -1$$

$$ \cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}}$$

$$ \cos\frac{\theta}{2} = \frac{\sqrt{2+2\cos\theta}}{2}$$

Using this, and starting from $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ yields


Therefore,  we can finish off this post, having derived Viète’s formula.

$$ \bbox[15px,border:1px solid black]{\frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\dots}$$


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