## 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

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

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

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The interior angle for each of these isosceles triangles is

$$\theta=\frac{2\pi}{n}$$

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

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

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

$$ Q_n= \cos\frac{\pi}{2^{n}}$$

Therefore

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

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

$$\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\hspace{20mm}\cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\hspace{20mm}\cos\frac{\pi}{16}=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}$$

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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