If `x+1/x=2 cos theta`, then the value of `x^n+1/x^n`

[lucaslugao]

Not really a mathematical proof.


x+1/x=2cos(θ)
x²+1=2x cos(θ)
x² - x (2cos(θ)) +1 = 0

x = [2cos(θ) ± √((-2cos(θ))²-4)]/2
x = cos(θ) ± √(4cos²(θ)-4)/2
x = cos(θ) ± √(cos²(θ)-1)

Since cos²(θ)-1 = -sin²(θ)

x = cos(θ) ± √(-sin²(θ))
x = cos(θ) ± i sin(θ)

Then we can put it in the exponential form

x = e^(±iθ)

Plugging it in the expression to be calculated

x^n + x^-n =
e^(±inθ) + e^(∓inθ) =
[cos(nθ) ± i sin(nθ)] + [cos(nθ) ∓ i sin(nθ)] =
2 cos(nθ)

For any value of θ