Tan Roots

all powers of x in the polynomial are even. so if tan(m1pi) is a solution, so is -tan(m1pi). -tan(m1pi)=tan(pi-m1pi).
so mi is of the form :.
m1=1-m2,.
m3=1-m4...
m1+m2=1,m3+m4=1,m5+m6=1,m7+m8=1;.
adding m1+m2+m3+m4+m5+m6+m7+m8=4

I show that the roots of the given polynomial are $tan(rπ/15)$ where $1≤r≤15$ and $gcd(r,15)=1$ .

Instead of proving that the roots if the polynomial are those $8$ numbers,I prove that the unique monic polynomial whose zeros are the $8$ numbers is $x^{8}-92x^{6}+134x^{4}-28x^{2}+1$

Note that if $\theta=rπ/15$ for $1≤r≤15$ and $gcd(r,15)=1$ ,then

$tan^{2}(5\theta)=3$ and $tan^{2}\theta≠3$ .

Now, I find the equation whose roots are $x=tan\theta$ .

$tan5\theta=\frac{Im(cos\theta+isin\theta)^{5}}{Re(cos\theta+isin\theta)^{5}}$

$=\frac{Im(1+itan\theta)^{5}}{Re(1+itan\theta)^{5}}$

$=\frac{Im(1+ix)^{5}}{Re(1+ix)^{5}}$

Expanding binomially, $(i+ix)^{5}=(1-10x^2+5x^4)+i(5x-10x^3+x^5)$

Hence, $tan5\theta=\frac{x(x^{4}-10x^{2}+5)}{5x^{4}-10x^{2}+1}$

Squaring, $\frac{x^{2}(x^{4}-10x^{2}+5)^{2}}{(5x^{4}-10x^{2}+1)^{2}}=3$

This reduces to $x^{10}-95x^8+410x^6-430x^4+85x^2-3=0$

or $(x^2-3)(x^8-92x^6+134x^4-28x^2+1)=0$

Since $x^2≠3$ , we have $(x^8-92x^6+134x^4-28x^2+1)=0$ .

Now the numbers relatively prime to $15$ are $1,2,4,7,8,11,13,14$

So the answer will be $\frac{1+2+4+7+8+11+13+14}{15}=\frac{60}{15}=4$

With z^4 - 92 z^3 + 134 z^2 - 28 z^2 + 1 = 0

x = +/- sqrt(z) = Tan [n Pi/ 15]

(1 + 2 + 4 + 7 + 14 + 13 + 11 + 8)/ 15 = 4