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Let $z = \cos(\frac{\pi k}{5}) + i\sin(\frac{\pi k}{5})$

Consider the imaginary part of $z^5$, and deduce that $x^4 - 3x^2 + 1 = 0$ has solutions:

$2\cos(\frac{\pi}{5}), ~2\cos(\frac{2\pi}{5}), ~2\cos(\frac{3\pi}{5}), ~2\cos(\frac{4\pi}{5})$

So, 'considering' the imaginary part of $z$, I considered the following to be true:

$5\cos^4\theta~\sin\theta - 10\cos^2\theta~\sin^3\theta + \sin^5\theta = 0$ where $\theta = \frac{\pi k}{5}$ since $z^5$ is real $\forall_k \in \mathbb{Z}$

How do I apply this to the polynomial from the question? Normally I would have just used the quadratic formula to solve it, but if I do that I get:

$x = \pm\sqrt{\frac{3 \pm \sqrt{5}}{2}}$

but I couldn't seem to relate the two sets of answers easily.

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If you assume $\sin\theta \neq 0$ you can divide through your trigonometric equation to get an equation of degree 4

Note that the given roots are in terms of $\cos k\theta$ so use $\sin^2\theta=1-\cos^2\theta$ to find a quartic in $\cos \theta$, and compare with the equation you have been given. Note the factor 2 in the given roots and make the appropriate substitution.

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    The primitive $n^{th}$ roots of unity are roots of the $n^{th}$ cyclotomic polynomial, which has integer coefficients. The roots of unity are also related to the field extensions which will contain the roots of a polynomial - there are problems in solving a polynomial if you don't have enough of the right roots of unity in your context. There is a huge literature on such things. These factors also make for problems which test the ability to manipulate trigonometric functions, related to polynomials with integer coefficients, where the integers seem to appear as if by magic.2012-06-12