Some Tricky Trigonometry

To solve this problem, we need to construct a cubic equation with the roots $\displaystyle \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{8\pi}{7}$ . Then we find the sum of the cube roots of that equation.

Starting from the $7^{\mathrm {th}}$ roots of unity, the roots of $x^7-1=0$ are $\displaystyle \left\{ e^{\frac{1}{7}2n\pi i} \space | \space n = 0, 1, \dots, 6\right\}$

By factoring $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$

We know that the roots of $x^6+x^5+x^4+x^3+x^2+x+1=0$ are $\displaystyle \left\{ e^{\frac{1}{7}2n\pi i} \space | \space n = 1, \dots, 6\right\}$

Deviding both sides by $x^3$ , we have $\displaystyle \left(x^3 + \frac{1}{x^3}\right) + \left(x^2 + \frac{1}{x^2}\right) +\left(x + \frac{1}{x}\right) = 0$

By the substitution $\displaystyle u = x + \frac{1}{x}$ , $\displaystyle u^2 - 2 = x^2 + \frac{1}{x^2}$ and $\displaystyle u^3-3u = x^3 + \frac{1}{x^3}$ ,

the equation reduces to $u^3+u^2-2u-1=0$ , and the roots

$\left\{ e^{\frac{1}{7}2n\pi i} + e^{-\frac{1}{7}2n\pi i} \space | \space n = 1, \dots, 6\right\} = \displaystyle \left\{ 2\cos\frac{2n\pi}{7} \space | \space n = 1, 2, 3\right\} = \left\{ 2\cos\frac{2\pi}{7}, 2\cos\frac{4\pi}{7}, 2\cos\frac{8\pi}{7} \right\}$

Note that $\left\{ \begin{matrix}\cos\frac{6\pi}{7} = \cos\frac{8\pi}{7} \\ \cos\frac{4\pi}{7} = \cos\frac{10\pi}{7} \\ \cos\frac{2\pi}{7} = \cos\frac{12\pi}{7} \end{matrix}\right.$ , and the Euler's identity $e^{i\theta}=\cos\theta + i\sin\theta$

The next step is to find the sum of the cube roots of $u^3+u^2-2u-1=0$

Let the roots of $u^3+u^2-2u-1=0$ be $a^3$ , $b^3$ , and $c^3$ .

By Vieta's formulae, $a^3+b^3+c^3=-1$ , and $a^3b^3+b^3c^3+a^3c^3=-2$ , and $a^3b^3c^3 = 1$

Substituding them in

$a^3+b^3+c^3 - 3abc=(a+b+c)^3-3(a+b+c)(ab+bc+ac)$

$a^3b^3+b^3c^3+a^3c^3 - 3 a^2b^2c^2 = (ab+bc+ac)^3-3abc(a+b+c)(ab+bc+ac)$

and letting $p=a+b+c$ and $q = ab+bc+ac$

we obtain the equations $\left\{ \begin{matrix}\displaystyle p^3 =3pq - 4 \\ \displaystyle q^3 = 3pq - 5 \end{matrix}\right.$

Multiplying them we get $(pq)^3-9(pq)^2+27pq-20=0$ , which, by completing the cubes, reduces to $(pq-3)^3+7=0$

So $pq = 3 - \sqrt[3]{7}$ , hence $p = \sqrt[3]{5-3\sqrt[3]{7}}$ and $q = \sqrt[3]{4 - 3\sqrt[3]{7} }$

Finally we have $\displaystyle \sqrt[3]{\cos\frac{2\pi}{7}} + \sqrt[3]{\cos\frac{4\pi}{7}} + \sqrt[3]{\cos\frac{8\pi}{7}} = \frac{p}{\sqrt[3]{2}} = \boxed{\displaystyle \sqrt[3]{\frac{1}{2}(5-3\sqrt[3]{7})}}$

link text