Polynomial in Cosine

Let $\omega = e^{2\pi i/7},$ so $\omega^7 = 1.$ Then $\cos \frac{2\pi} 7 = \frac12 (\omega + 1/\omega),$ by Euler's formula . Then $\begin{aligned} 2\cos^3 \frac{2\pi}7 + \cos^2 \frac{2\pi}7 - \cos \frac{2\pi}7 &= \frac28 (\omega + 1/\omega)^3 + \frac14 (\omega + 1/\omega)^2 - \frac12(\omega + 1/\omega) \\ &= \frac14(\omega^3 + 3\omega + 3/\omega + 1/\omega^3 + \omega^2 + 2 + 1/\omega^2 - 2\omega - 2/\omega) \\ &= \frac14 + \frac14(\omega^3 + \omega^2 + \omega + 1 + 1/\omega + 1/\omega^2 + 1/\omega^3) \\ &= \frac14 + \frac1{4\omega^3} ( \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) \\ &= \frac14 + \frac1{4\omega^3} \frac{1-\omega^7}{1-\omega} \\ &= \frac14 = \fbox{0.25}. \end{aligned}$

This implies that $\cos \frac{2\pi}7$ is a root of the polynomial $8x^3+4x^2-4x-1,$ which can be easily checked to be irreducible over the rational numbers (e.g. by the rational root theorem ), so its degree is $3,$ which implies that it is not constructible .

$\begin{aligned} X & = 2\cos^3 \frac {2\pi}7 + \cos^2 \frac {2\pi}7 - \cos \frac {2\pi}7 \\ & = \frac 12 \left(4\cos^3 \frac {2\pi}7 - 3\cos \frac {2\pi}7\right) + \frac 32 \cos \frac {2\pi}7 + \frac 12 \left(\cos \frac {4\pi}7 + 1\right) - \cos \frac {2\pi}7 \\ & = \frac 12 \left(\color{#3D99F6} \cos \frac {2\pi}7 + \cos \frac {4\pi}7 + \cos \frac {6\pi}7 \right) + \frac 12 & \small \color{#3D99F6} \text{See proof} \\ & = \frac 12 \left(\color{#3D99F6} - \frac 12 \right) + \frac 12 \\ & = \frac 14 = \boxed{0.25} \end{aligned}$

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Proof for $\color{#3D99F6} \displaystyle \sum_{k=1}^n \cos \left(\frac {2k}{2n+1}\right) = - \frac 12$

Disclaimer: I cheated with a calculator and I reverse-engineered the answer.

We start with a famous trigonometric identity, $\cos\frac{2\pi}7 + \cos\frac{4\pi}7 + \cos\frac{6\pi}7 = -\frac12 \quad \Leftrightarrow \quad 2\cos\frac{6\pi}7 + 2\cos\frac{4\pi}7 + 2\cos\frac{2\pi}7 + 1 = 0.$

Let $x = \cos\frac{2\pi}7$ , using double angle formula and triple angle formula, we have $\cos\frac{4\pi}7 = 2x^2 - 1$ and $\cos\frac{6\pi}7 = 4x^3 - 3x$ . And the equation above can be written as $2(4x^3 - 3x) + 2(2x^2 - 1) + 2x + 1 = 0 \quad \Leftrightarrow \quad 4(2x^3 + x^2 - x) = 1.$ Hence, the answer is simply $2x^3 + x^2 - x = \frac14 = \boxed{0.25}$ .

Now I'm wondering how does this relate to the author's last statement:

This equation is key step to show that it is impossible to draw the regular heptagon with straightedge-and-compass.

Direct evaluation: $2 \cos ^3\left(\frac{2 \pi }{7}\right)+\cos ^2\left(\frac{2 \pi }{7}\right)-\cos \left(\frac{2 \pi }{7}\right) \Rightarrow \frac14$ .

$\frac{1}{2}+\frac{1}{4} i e^{-\frac{1}{14} (3 i \pi )}-\frac{1}{4} i e^{\frac{3 i \pi }{14}}-\frac{1}{4} e^{-\frac{1}{7} (3 i \pi )}-\frac{1}{4} e^{\frac{3 i \pi }{7}}-\frac{1}{4} i e^{-\frac{1}{14} (9 i \pi )}+\frac{1}{4} i e^{\frac{9 i \pi }{14}} \Rightarrow \frac{1}{4} \left(2-\sqrt[7]{-1}+(-1)^{2/7}-(-1)^{3/7}+(-1)^{4/7}-(-1)^{5/7}+(-1)^{6/7}\right) \Rightarrow \frac14\,(2-1) \Rightarrow \frac14$