<s>[INST] Cyclotomic polynomials and products of cosines problem I've run into an inconsistency I can't figure out while trying to find products of the cosine of various roots of unity.
For example: $\cos(\frac{2\pi}{5})\cdot cos(\frac{4\pi}{5}) \cdot cos(\frac{6\pi}{5}) \cdot cos(\frac{10\pi}{5})$ Multiply everything by $\frac{\sin(\frac{2\pi}{5})}{sin(\frac{2\pi}{5})}$ and apply the trig rules and you get $\frac{1}{16}$
(which is correct)
These are all part of the 5th roots of unity which are solutions to $z^5 - 1 = 0$ and all of the form $\cos(x) + i\sin(x)$  So you can substitute that in expand and replace $\sin(x)^2 = 1 - cos(x)^2$  and split by the real and imaginary parts. For the reals you get $16\cos(x)^5 -10\cos(x))^3 + 5\cos(x) - 1 = 0$ and by the vieta formula this also implies that product of the cosine of the 5 roots is $\frac{1}{16}$  One root is 1 so this is consistent with first method.
But then I was looking at $\Phi_5 = z^4 + z^3 + z^2 + z + 1$  The four root besides one are roots of this polynomial and I should be able to do the same substitution from above. 
let c = cos(x) and s =sin(x)
$(c + is)^4 + (c + is)^3 + (c + is)^2 + c + is + 1 = 0$
$(c^4 + 4ic^3s -6c^2s^2 -4ics^3 + s^4) + (c^3 + 3ic^2s - 3cs^2 - is^3) + (c^2 + 2ics - s^2) + (c + is) + 1 = 0$
For the real part:
$c^4 -6c^2s^2  + s^4 + c^3 - 3cs^2 + c^2 - s^2 + c + 1 = 0$ 
Then substituting $s^2 = 1 - c^2$ I get
$8\cos(x)^4 + 4\cos(x)^3 - 6\cos(x)^2 - 2\cos(x) + 1 = 0$ which implies that the product of the cosines is $\frac{1}{8}$
What's going on?
 [/INST] Thanks that gave me the idea to figure out what went wrong. The key is its necessary to solve both the imaginary and real parts of the derived equation and throw out solutions that don't handle both.  Its interesting what extra solutions did show up and how this method "eliminated" the duplicate roots.
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