Polynomial with known root tangent 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 $\tan^2 \theta=\dfrac{1-\cos^2\theta}{\cos^2\theta}$ , so we let $t=\tan^2\left(\dfrac{2 \pi}{15}\right)$ :

$t=\dfrac{1-\frac{c^2}{4}}{\frac{c^2}{4}} \implies c=\dfrac{2}{\sqrt{t+1}}$

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

$t^4-92t^3+134t^2-28t+1=0$

We compare that with the given form, and the final answer is $92+134+28=\boxed{254}$ .

Tan^2 (12 d), Tan^2 (24 d), Tan^2 (48 d), Tan^2 (96 d) are roots of x^4 - 92 x^3 + 134 x^2 - 28 x + 1 = 0.

The polynomial is a sole equation of positive integers a, b and c. There is no simple surd (A + B Sqrt (C))/ D seemingly for describing exact value of Tan^2 (24 d) but perhaps another form. All because of integer coefficients.

a + b + c = 92 + 134 + 28 = 254