Polynomial with known root sine squared

We start with the cyclotomic polynomial $\Phi_{15}(x)=x^8-x^7+x^5-x^4+x^3-x+1$ . Then we know that $c=x+\dfrac{1}{x}=2\cos\left(\dfrac{2 \pi}{15}\right)$ . We can divide the polynomial by $x^4$ and then rearrange to obtain:

$x^4+\dfrac{1}{x^4}-\left(x^3+\dfrac{1}{x^3}\right)+x+\dfrac{1}{x}-1=0$

Now, do some algebra to express that only in terms of $c=x+\dfrac{1}{x}$ :

$\left(x+\dfrac{1}{x}\right)^4-\left(x+\dfrac{1}{x}\right)^3-4\left(x+\dfrac{1}{x}\right)^2+4\left(x+\dfrac{1}{x}\right)+1=0 \\ c^4-c^3-4c^2+4c+1=0$

That is the minimal polynomial of $2\cos\left(\dfrac{2 \pi}{15}\right)$ , finally, use the identity $\sin^2 \theta=1-\cos^2 \theta$ , so we let $s=\sin^2\left(\dfrac{2 \pi}{15}\right)$ :

$s=1-\dfrac{c^2}{4} \implies c=2\sqrt{1-s}$

Substitute that into the equation in $c$ and after some algebra we end up with:

$256s^4-448s^3+224s^2-32s+1=0$

We compare that with the given form, and the final answer is $256+448+224+32=\boxed{960}$ .