Valore esatto di cos(2π/5)

Exercise for high school students who have studied complexes and are wondering what they can be used for. Let’s show that

cos(2π/5)=(-1+√5)/4. (1)

Let w=cos(2π/5)+isin(2π/5) and x=Re(w). We want to show that x= (-1+√5)/4. Let’s observe that:

x=(w+w⁻¹)/2 (2)

where w⁻¹ is the multiplicative inverse of w. To see this, let’s first recall that if z=a+ib is a complex number z⁻¹=(a-ib)/(a²+b²) and therefore w⁻¹=(cos(2π/5)-isin(2π/5))/ 1= cos(2π/5)-isin(2π/5) from which immediately follows (2) (where in the first equality we used the well-known fact that cos²(2π/5)+sin²(2π/5)=1). Note that w⁻¹ is precisely the conjugate of w. Now let’s observe that w is the root of the complex equation z⁵=1. In fact:

w⁵=cos(2π)+isin(2π)=1

then:

(w-1)(w⁴+w³+w²+w+1)=0

it follows, since w≠1, that (we are using the fact that ℂ is an integral domain)

w⁴+w³+w²+w+1 = 0

Now, since w satisfies this equation (with coefficients in Z) its conjugate also satisfies it, i.e. w⁻¹, it follows:

w⁻⁵+w⁻⁴+w⁻³+w⁻²+w⁻¹+1=0.

Now:

(2x)²+(2x)-1 =
(w+w⁻¹)²+(w+w⁻¹)-1=
w²+w⁻²+2+w+w⁻¹-1=
w²+1/w²+w+1/w + 1 =
0/w²=0

It follows that , being x > 0:

x=(-2+2√5)/8=(-1+√5)/4

the assertion follow.

Corollary:

cos(4π/5)=(-1-√5)/4

Proof:

It follows from formula (1) and from the formula cos(2α)=2(cos(α))²-1.