Threatening Trigonometry

With the identity $\cos(\theta) = -cos(\pi-\theta)$ we can rewrite the expression as:

$\cos^4 \dfrac{2\pi}{7}+\cos^4 \dfrac{4\pi}{7}+\cos^4 \dfrac{6\pi}{7}$

Now, we are going to find an equation with roots $\cos \dfrac{2\pi}{7}$ , $\cos \dfrac{4\pi}{7}$ and $\cos \dfrac{6\pi}{7}$ . Let $w^7=1$ , so $(w-1)(w^6+w^5+w^4+w^3+w^2+w+1)=0$ .

Let's focus only on the second factor. Divide both sides by $w^3$ :

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

Factorize and simplify:

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

Now, since $w^7=1$ , lets choose the first primitive root, so, $w=e^\frac{2\pi}{7}$ and $w+\dfrac{1}{w}=2\cos\dfrac{2\pi}{7}$ . But, if we choose the second root, $w+\dfrac{1}{w}=2\cos \dfrac{4\pi}{7}$ and if we choose the third root, $w+\dfrac{1}{w}=2\cos \dfrac{6\pi}{7}$ . If we choose the other roots, we obtain the same results. So, let $x=w+\dfrac{1}{w}$ :

$x^3+x^2-2x-1=0$

That equation has roots $2 \cos \dfrac{2\pi}{7}$ , $2 \cos \dfrac{4\pi}{7}$ and $2 \cos \dfrac{6\pi}{7}$ , so let $x=2y$ :

$8y^3+4y^2-4y-1=0 \Longrightarrow y^3+\dfrac{1}{2}y^2-\dfrac{1}{2}y-\dfrac{1}{8}=0$ .

This is the equation we wanted. Finally, we can apply Newton's Sums to obtain the initial expression:

$P_n=\cos^n \dfrac{2\pi}{7}+\cos^n \dfrac{4\pi}{7}+\cos^n \dfrac{6\pi}{7} \\ P_1=-\dfrac{1}{2} \\ P_2=-\dfrac{1}{2}P_1-2\left(-\dfrac{1}{2}\right)=\dfrac{5}{4} \\ P_3=-\dfrac{1}{2}P_2-\left(-\dfrac{1}{2}\right)P_1-3\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2} \\ P_4=-\dfrac{1}{2}P_3-\left(-\dfrac{1}{2}\right)P_2-\left(-\dfrac{1}{8}\right)P_1=\boxed{\dfrac{13}{16}}$

So, $a=13$ , $b=16$ and $a+b=\boxed{29}$ .