Show that cos(π/5) equals half the golden ratio

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Homework Statement


Show that cos(π/5) = λ/2, where λ = (1 +√5)/2 (the Golden Ratio).

Homework Equations


[Hint: As cos 5θ = 1, where θ = 2π/5, we see from De Moivre’s theorem
that P(cos θ) = 0 for some polynomial P of degree five. Now observe that
P(z) = (1 − z)Q(z)2 for some quadratic polynomial Q.]

The Attempt at a Solution



Not sure how!
from P(cos θ) = 0
there are 5 solutions for p(z) and one solution for θ. but then only one of the 6 multiplying factors needs to be zero right?
i.e. ## (z - a)(z - b)(z - c)(z - d)(z - e)(Cos θ) = 0 ## only one of them needs to be zero because anything multiplied zero is also zero.
 
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PcumP_Ravenclaw said:

Homework Statement


Show that cos(π/5) = λ/2, where λ = (1 +√5)/2 (the Golden Ratio).

Homework Equations


[Hint: As cos 5θ = 1, where θ = 2π/5, we see from De Moivre’s theorem
that P(cos θ) = 0 for some polynomial P of degree five. Now observe that
P(z) = (1 − z)Q(z)2 for some quadratic polynomial Q.]
Is the last factor supposed to be Q(z)2?
PcumP_Ravenclaw said:

The Attempt at a Solution



Not sure how!
from P(cos θ) = 0
there are 5 solutions for p(z) and one solution for θ. but then only one of the 6 multiplying factors needs to be zero right?
No. From P(z) = 0, there are 5 solutions, but from P(cos θ) = 0, θ doesn't need to be zero and cos(θ) doesn't need to be zero.
PcumP_Ravenclaw said:
i.e. ## (z - a)(z - b)(z - c)(z - d)(z - e)(Cos θ) = 0 ## only one of them needs to be zero because anything multiplied zero is also zero.