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

  1. #1
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    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.
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  2. #2
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    Let z=\mathrm e^{\mathrm i\pi/5}. Show that x=z+z^{-1} and note that z^2\neq 1.

    Then show that x^2-2=z^2+z^{-2} and that (x^2-2)^2-2=z^4+z^{-4}.

    Deduce that x^4-3x^2+1=z^4+z^2+1+z^{-2}+z^{-4}=\frac{z^{10}-1}{z^4(z^2-1)}.

    Consider the value of z^{10} and chillax.
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  3. #3
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    Talking

    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
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