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.


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.

  • $\begingroup$ OK, so I got the equation that shows they are indeed roots; however, I feel like I have missed the point of the exercise somehow. What is it trying to show? How did you immediately know that you should divide by $\sin\theta$ etc.? Is this just a contrived question/polynomial that works out nicely, or what is the relationship that I am missing? I can do the division and the substituting that you told me to do :D but I don't know if I would know what to do given a similar problem. $\endgroup$ – stariz77 Jun 12 '12 at 7:24
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 12 '12 at 9:35

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