Computing roots of unity

Hi,

the problem states.

Express cos2π/5 and sin2π/5 in terms of roots of real numbers.

now I know that (cos2pi/5+isin2pi/5)^5=cos2pi+isin2pi=1=… do i go from here?

Note that (cos 2π/5 + i sin 2π/5)^5 = cos 2π + i sin 2π = 1.
So, z = cos 2π/5 + i sin 2π/5 satisfies z^5 = 1.

To solve this (and find cos 2π/5 + i sin 2π/5 in terms of radicals):
z^5 - 1 = 0
==> (z - 1)(z^4 + z^3 + z^2 + z + 1) = 0
==> z^4 + z^3 + z^2 + z + 1 = 0, since we want z ≠ 1.

Now comes a trick; divide through by z^2:
z^2 + z + 1 + 1/z + 1/z^2 = 0
==> (z^2 + 1/z^2) + (z + 1/z) + 1 = 0
==> (z^2 + 2 + 1/z^2) - 2 + (z + 1/z) + 1 = 0
==> (z + 1/z)^2 + (z + 1/z) - 1 = 0.

Let w = z + 1/z.
==> w^2 + w - 1 = 0.

Solving for w yields
w = (-1 ± √5)/2

So, z + 1/z = (-1 ± √5)/2
==> 2z^2 + (1 ± √5)z + 2 = 0
==> z = [-(1 ± √5) + √(-10 ± 2√5)] / 4 or [-(1 ± √5) - √(-10 ± 2√5)] / 4
==> z = [-(1 ± √5) + i √(10 ∓ 2√5)] / 4 or [-(1 ± √5) - i√(10 ∓ 2√5)] / 4

Remembering that z = cos 2π/5 + i sin 2π/5 has both positive real and imaginary parts, we conclude that z = [-(1 - √5) + i √(10 + 2√5)] / 4.

Equating real and imaginary parts yields
cos 2π/5 = (√5 - 1)/4 and sin 2π/5 = √(10 + 2√5)/4.

I hope this helps!