$\sin (5\theta )=16{\sin }^5\theta -20{\sin }^3\theta +5\sin \theta$

Now letting $\theta =2\pi /5$, we get a polynomial in $\sin (2\pi /5$ (which is what we want) equal to 0 (because $\sin (5\theta )$ is cunningly zero). Hopefully you can see how to solve this (it reduces to a quadratic very easily), and you can get $\cos (2\pi /5)$ from this, and the other angles.

If you want this explaining any more, let us know!

Vicky

Although I'm not sure of the mathematical validity of how I 'solved' the quadratic.

You have:

$\sin (5\theta )=16{\sin }^5\theta -20{\sin }^3\theta +5\sin \theta$

And setting $\theta =2\pi /5$ you get:

$16{\sin }^5(2\pi /5)-20{\sin }^3(2\pi /5)+5\sin (2\pi /5)=0$

Therefore:

$\sin (2\pi /5)[16{\sin }^4(2\pi /5)-20{\sin }^2(2\pi /5)+5]=0$

Now, must you say that either

$\sin (2\pi /5)=0$

or

$16{\sin }^4(2\pi /5)-20{\sin }^2(2\pi /5)+5=0$

Or can you simply divide both sides by $\sin (2\pi /5)$ as we 'know' already that $\sin (2\pi /5)\ne 0$?

I guess I'm confused between constants and variables.