# A Beautiful Expression for Pi

## Proving a surprising and pleasing mathematical result

--

Today I am going to prove a beautiful and somewhat surprising expression for π. The expression that I will prove is the following:

where the denominator is an infinite product of nested surds.

Step 1 of my proof will establish an important trigonometric identity, which we will use in Step 2 by taking limits and applying the result to π.

# Proving an important trigonometric identity

Our expression for π can be derived from the following trigonometric identity:

for all positive integers *n *and where ⍺ is not an integer multiple of π.

We are going to use classic induction to prove this. Remember that you can prove a result for all positive integers *n* if and only if you can prove it is true for *n = 1* (the induction start), and you can prove that if it is true for *n = k* then it is true for *n = k + 1 *(the induction step)*.*

To prove the induction start, let’s use the standard trigonometric identity sin(θ+ϕ) = sinθcosϕ + cosθsinϕ. If θ = ϕ = ⍺/2, we have sin⍺ = 2sin(⍺/2)cos(⍺/2). Dividing both sides by cos(⍺/2) gives the result for *n = *1*.*

Now let’s assume the result is true for *n = k. *We can use the same standard trigonometric identity to derive the result for *n = k + 1*, as follows:

Hence the result is true for all positive integers *n*.

# Using this to prove our beautiful expression for π

Now we make the observation that as an angle θ approaches zero, the value of (sinθ)/θ approaches 1. From this, with a little simple manipulation, we can say that

So applying limits to both sides of our trigonometric identity from the last section, we can conclude the following:

Now we are going to use another well known trigonometric identity, which is that cos(θ+ϕ) = cosθcosϕ -sinθsinϕ. Again, with θ = ϕ = ⍺/2, we can say that cos⍺ = cos²(⍺/2)-sin²(⍺/2) =…