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?

$$z=\cos(2k\pi/5)+i\sin(2k\pi/5)$$
($k\in\{1,2,3,4\}$) are the zeros of $z^4+z^3+z^2+z+1$. Then
$w=z+1/z=2\cos(2k\pi/5)$ and
$$0=z^2+z+1+z^{-1}+z^{-2}=(z+z^{-1})^2+(z+z^{-1})-1=w^2+w-1.$$
Each $\cos(2k\pi/5)$ is a zero of
$$4x^2+2x-1$$
and so a zero of
$$(4x^2+2x-1)^2=16x^4+16x^3-4x^2-4x+1.$$
I suspect this is the equation you were after.
[content added in edit]
Your equation is
$$8y^4+4y^3-6y^2-2y+1=0.$$
This factors as
$$(4y^2+2y-1)(2y^2-1)$$
so has zeros $\cos(2\pi/5)$, $\cos(4\pi/5)$, $1/\sqrt2$ and $-1/\sqrt2$.
From this we correctly deduce
$$\left(\cos\frac{2\pi}5\right)\left(\cos\frac{4\pi}5\right)
\left(\frac1{\sqrt2}\right)\left(-\frac1{\sqrt2}\right)=\frac18.$$
