Thread: show that x^4 - 3x^2 + 1 = 0 when x is given as 2cos(π/5)

let x = 2cos(π/5)

use complex numbers to show that

x^4 - 3x^2 + 1 = 0

I'm stumped completely

I tried substituting the value in and using trig identities.. got a big mess. I tried treating the x as the x part of a complx number... still more mess.

Anything to at least lead me in the right direction... I'm sure it's staring me in the face.

Let . Show that and note that .

Then show that and that .

Deduce that .

Consider the value of and chillax.

Using Euler's Formula

e^(5iθ) = (cos(θ) + i sin(θ))^5

so

sin(5θ) = Im[(cos(θ) + i sin(θ))^5]
= sin(θ)[5 cos^4(θ) - 10 cos²(θ)sin²(θ) + sin^4(θ)]
= sin(θ)[5 cos^4(θ) - 10 cos²(θ)(1 - cos²(θ)) + (1 - cos²(θ))²
= sin(θ)[16 cos^4(θ) - 12 cos²(θ) + 1]

Let x = 2 cos(π/5); then cos(π/5) = x/2. Now, evaluate the above result at θ = π/5:

0 = sin(5π) = sin(π/5)[[16 cos^4(π/5) - 12 cos²(π/5) + 1]

Replacing cos(π/5) by x/2 and reducing, we get

0 = sin(π/5)[[x^4 - 3x² + 1]

But sin(π/5) ≠ 0 so we must have

x^4 - 3x² + 1 = 0