Simplifying $\cos\left(\frac{2 \pi}{21}\right) + \cos\left(\frac{8 \pi}{21}\right) + \cos\left(\frac{10 \pi}{21}\right)$

Question

Given $$ S = \cos\left(\frac{2 \pi}{21}\right) + \cos\left(\frac{8 \pi}{21}\right) + \cos\left(\frac{10 \pi}{21}\right)$$ then what are some, or at least one, method to reduce this sum? According to Wolfram Alpha/Mathematica $$S = \frac{1 + \sqrt{21}}{4}.$$

Attempt

By using \begin{align} (a + b + c)^3 &= 6 a b c + 3 (a + b + c)(a^2 + b^2 + c^2) - 2 (a^3 + b^3 + c^3) \\ a^2 + b^2 + c^2 &= (a + b + c)^2 - 2 (ab + bc + ca) \\ \cos^3 \theta &= \frac{1}{4} \, ( 3 \cos\theta + \cos(3 \theta)) \end{align} or $$ (a + b + c)^3 = (a^3 + b^3 + c^3) - 3 (a + b + c)(ab + bc + ca) - 3 a b c$$ then $$(\cos x + \cos y + \cos z)^3 - \frac{3}{4} (\cos x + \cos y + \cos z) = \frac{1}{4} (\cos 3x + \cos 3y + \cos 3z) - 3 (\cos x + \cos y + \cos z) (\cos x \cos y + \cos y \cos z + \cos z \cos x) - 3 \cos x \cos y \cos z.$$ If $$ x = \frac{2 \pi}{21} \quad y = \frac{8 \pi}{21} \quad z = \frac{10 \pi}{21}$$ then $$S^3 - \frac{3}{4} \, S = \frac{x}{4} - 3 S \, P - 3 \, \cos\left(\frac{2 \pi}{21}\right) \cos\left(\frac{8 \pi}{21}\right) \cos\left(\frac{10 \pi}{21}\right), $$ where \begin{align} x &= \cos\left(\frac{2 \pi}{7}\right) + \cos\left(\frac{8 \pi}{7}\right) + \cos\left(\frac{10 \pi}{7}\right) \\ &= - \left(\cos\left(\frac{\pi}{7}\right) - \cos\left(\frac{2 \pi}{7}\right) + \cos\left(\frac{3 \pi}{7}\right) \right) \\ P &= \cos\left(\frac{2 \pi}{21}\right) \cos\left(\frac{8 \pi}{21}\right) + \cos\left(\frac{8 \pi}{21}\right) \cos\left(\frac{10 \pi}{21}\right) + \cos\left(\frac{10 \pi}{21}\right) \cos\left(\frac{2 \pi}{21}\right). \end{align}

Since, \begin{align} P &= \cos\left(\frac{2 \pi}{21}\right) \cos\left(\frac{8 \pi}{21}\right) + \cos\left(\frac{8 \pi}{21}\right) \cos\left(\frac{10 \pi}{21}\right) + \cos\left(\frac{10 \pi}{21}\right) \cos\left(\frac{2 \pi}{21}\right) \\ &= \frac{1}{2} \, \left[ \left(\cos\left(\frac{2 \pi}{7}\right) + \cos\left(\frac{4 \pi}{7}\right) + \cos\left(\frac{6 \pi}{7}\right) \right) + \cos\left(\frac{2 \pi}{21}\right) + \cos\left(\frac{8 \pi}{21}\right) + \cos\left(\frac{10 \pi}{21}\right) \right] \\ &= \frac{1}{2} \, \left[ S - \left(\cos\left(\frac{\pi}{7}\right) - \cos\left(\frac{2 \pi}{7}\right) + \cos\left(\frac{3 \pi}{7}\right) \right) \right] \\ &= \frac{1}{2} \, (S + x), \end{align} then $$S^3 + \frac{3}{2} \, S^2 - \frac{3}{4} \, S = \frac{x}{4} - \frac{3 \, x \, S}{4} - 3 \, \cos\left(\frac{2 \pi}{21}\right) \cos\left(\frac{8 \pi}{21}\right) \cos\left(\frac{10 \pi}{21}\right).$$

This seems to be trading one complication for another.

Note The original problem was related to a reduction of $$x = \cos\left(\frac{\pi}{7}\right) + \cos\left(\frac{4 \pi}{7}\right) + \cos\left(\frac{5 \pi}{7}\right)$$ but was in error since, in my notes, I had made a reduction of $\frac{2}{3}$ to each term.

Answer 1

Let $a = \frac\pi{21}$ and

$$ S = \cos2a + \cos 8a+\cos10a, \>\>\>\>\> T= \cos4a + \cos 16a+\cos20a$$

Use the trig identity $2\cos x\cos y = \cos(x-y)+\cos(x+y)$ to get $$2\cos\frac\pi3(\cos\frac\pi7+\cos\frac{3\pi}7+\cos\frac{5\pi}7)=S+T$$

Knowing $\cos\frac\pi7+\cos\frac{3\pi}7+\cos\frac{5\pi}7=\frac12$, we have

$$S+T= \frac12\tag 1$$ Use the trig identity again to evaluate $$\begin{align} 2ST &=2( \cos2a + \cos 8a+\cos10a)(\cos4a + \cos 16a+\cos20a) \\ & = S+T + 3(\cos6a+\cos12a+\cos14a+\cos18a) \\ & = \frac12 +3\cos\frac{2\pi}3- 3(\cos\frac\pi7+\cos\frac{3\pi}7+\cos\frac{5\pi}7) \\ & = \frac12 -\frac32 -\frac32 = -\frac52 \\ \end{align}$$

So,

$$ST = -\frac54\tag 2$$

From (1) and (2), $S$ and $T$ satisfy the quadratic equation,

$$x^2-\frac12x-\frac54=0$$

which yields

$$S = \frac{1+\sqrt{21}}4$$

Answer 2

I'm pretty sure the problem was intended to be $$ x = \cos\left(\frac{\pi}{7}\right) + \cos\left(\frac{3 \pi}{7}\right) + \cos\left(\frac{5 \pi}{7}\right)$$ because this one really does simplify.

Take $$ \omega = e^{2 \pi i / 7} $$ so that $\omega^7 = 1$ and $\omega \neq 1$ and $$ 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0. $$ Well, as pointed out by Gauss, take $$ \alpha = \omega + \omega^2 + \omega^4 $$ so that $$ \alpha + \bar{\alpha } = -1. $$ Since $\omega^8 = \omega,$ we find $$ \alpha^2 = \omega^2 + \omega^4 + \omega^8 + 2 \omega^3 + 2 \omega^5 + 2 \omega^6, $$ $$ \alpha^2 + \alpha = 2( \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 ) $$ $$ \alpha^2 + \alpha + 2 = 0 $$ so that $$ \alpha = \frac{-1 \pm i \sqrt 7}{2} $$ Meanwhile, $$ \alpha = \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} \right)+ i \left( \sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7} \right)$$

$$ \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} =- \frac{1}{2}$$ $$ - \cos \frac{5 \pi}{7} - \cos \frac{3 \pi}{7} - \cos \frac{ \pi}{7} = - \frac{1}{2}$$ $$ \cos \frac{5 \pi}{7} + \cos \frac{3 \pi}{7} + \cos \frac{ \pi}{7} = \frac{1}{2}$$

Answer 3

About your new problem with the 21's.

Here is a modern presentation, two chapters about the Gauss method, book by Cox.

If $\omega$ is a primitive 21st root of unity, and we take $$ \alpha = \omega + \omega^4 + \omega^{16} \; , \; $$ we find that $$ \alpha^4 - \alpha^3 - \alpha^2 - 2 \alpha + 4 = 0. $$ See page 202 in Reuschle(1875). This polynomial is irreducible over the rationals, but factors as $$ \left( x^2 - \left( \frac{1 + \sqrt{21}}{2} \right) x + 2 \right) \left( x^2 - \left( \frac{1 - \sqrt{21}}{2} \right) x + 2 \right) $$

Answer 4

COMMENT.- I apologize for the attached figure to ilustrated a comment above.