What happened to $3\pi/9$ ?

Let $w_1=e^{\pi i/9},$ $w_2=e^{3\pi i/9},$ $w_3=e^{5\pi i/9},$ and $w_4=e^{7\pi i/9}.$ Then the numbers $w_1, w_2, w_3, w_4, \overline{w_1}, \overline{w_2}, \overline{w_3}, \overline{w_4}$ , and -1 are the roots of the polynomial $x^9+1.$ Therefore, $\begin{aligned} x^9+1=(x+1)(x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1)=\\ =(x+1)(x-w_1)(x-\overline{w_1})(x-w_2)(x-\overline{w_2})(x-w_3)(x-\overline{w_3})(x-w_4)(x-\overline{w_4})\end{aligned}.$ Dividing both sides by $x+1$ and taking into consideration that $(x-w_j)(x-\overline{w_j})=x^2-2\cos{ \frac{(2j-1)\pi} {9}} x+1,$ for $j\in \{1,2,3,4\},$ we obtain that $x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1=\\ =\prod_{j=1}^4 (x^2-2\cos{\frac{(2j-1)\pi} {9}} x+1).$ Making the coefficients of $x$ in this equality equal, and using that $\cos(3\pi /9)=\cos(\pi/3)=\frac{1}{2},$ we get the equality $\cos(\frac{\pi}{9})+\cos(\frac{5\pi}{9})+\cos(\frac{7\pi}{9})=0. \:\:\:\:(*)$

In a similar way, making the coefficients of $x^2$ and $x^4$ equal, respectively, and considering again that $\cos(3\pi /9)=\frac{1}{2},$ we obtain the equalities

$\cos(\frac{\pi}{9})\cos(\frac{5\pi}{9})+\cos(\frac{7\pi}{9})\cos(\frac{5\pi}{9})+\cos(\frac{\pi}{9})\cos(\frac{7\pi}{9})=-\frac{3}{4} \:\:\:\:(**)$ and $\cos(\frac{\pi}{9})\cos(\frac{5\pi}{9})\cos(\frac{7\pi}{9})=\frac{1}{8}.\:\:\:\:(***)$ Now, using $(*), (**)$ and $(***)$ and Vieta's Formulas, we get that $f(x)= 8x^3-6x-1.$ Therefore, $f(0)+f(1)+f(10)=\boxed{7939}.$

Let $a = \cos \dfrac \pi 9$ , $b = \cos \dfrac {5\pi} 9$ and $c = \cos \dfrac {7\pi} 9$ . Then, we have:

$\begin{aligned} f(x) & = 8 \left(x-\cos \frac \pi 9 \right) \left(x-\cos \frac {5\pi} 9 \right) \left(x-\cos \frac {7\pi} 9 \right) \\ & = 8 \left(x-a \right) \left(x-b \right) \left(x-c \right) \\ & = 8\left(x^3 - (\color{#3D99F6}{a+b+c})x^2 + (\color{#3D99F6}{ab+bc+ca})x -\color{#3D99F6}{abc} \right) \quad \quad \small \color{#3D99F6}{\text{See Note.}} \\ & = 8\left(x^3 - (\color{#3D99F6}{0})x^2 + \left(\color{#3D99F6}{-\frac 34}\right)x -\color{#3D99F6}{\frac 18} \right) \\ & = 8x^3 - 6x -1 \end{aligned}$

$\begin{aligned} \implies f(0) + f(1) + f(10) & = -1 + 1 + 8000 - 60 - 1 = \boxed{7939} \end{aligned}$

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$\color{#3D99F6}{\text{Note:}}$

$\begin{aligned} \cos \frac \pi 9 + \cos \frac {3\pi} 9 + \cos \frac {5\pi} 9 + \cos \frac {7\pi} 9 & = \frac 12 \\ \implies a + \frac 12 + b + c & = \frac 12 \\ \color{#3D99F6}{\implies a + b + c} & \color{#3D99F6}{= 0} \end{aligned}$

$\begin{aligned} \implies a^2+b^2+c^2 & = -2(ab+bc+ca) \\ \cos^2 \frac \pi 9 + \cos^2 \frac {5\pi} 9 + \cos^2 \frac {7\pi} 9 & = \frac 12 \left(\cos \frac {2\pi} 9 + \cos \frac {10\pi} 9 + \cos \frac {14\pi} 9+3 \right) \\ & = \frac 12 \left(-\cos \frac {7\pi} 9 - \cos \frac {1\pi} 9 - \cos \frac {5\pi} 9+3 \right) \\ & = \frac 12 \left(-0+3 \right) = \frac 32 \\ \color{#3D99F6}{\implies ab+bc+ca} & \color{#3D99F6}{ = - \frac 34} \end{aligned}$

$\begin{aligned} abc & = \cos \frac \pi 9 \cos \frac {5\pi} 9 \cos \frac {7\pi} 9 = \cos \frac \pi 9 \cos \frac {4\pi} 9 \cos \frac {2\pi} 9 \\ & = \frac {\sin \frac \pi 9 \cos \frac \pi 9 \cos \frac {2\pi} 9 \cos \frac {4\pi} 9}{\sin \frac \pi 9} = \frac {\sin \frac {2\pi} 9 \cos \frac {2\pi} 9 \cos \frac {4\pi} 9}{2\sin \frac \pi 9} \\ & = \frac {\sin \frac {4\pi} 9 \cos \frac {4\pi} 9}{4\sin \frac \pi 9} = \frac {\sin \frac {8\pi} 9}{8\sin \frac \pi 9} = \frac {\sin \frac \pi 9}{8\sin \frac \pi 9} \\ \color{#3D99F6}{\implies abc} & \color{#3D99F6}{ = \frac 18} \end{aligned}$

$f(x) = 8\left(x-\cos\dfrac{\pi}{9}\right)\left(x-\cos\dfrac{5\pi}{9}\right)\left(x-\cos\dfrac{7\pi}{9}\right)\\ =8\left(x-\cos20^{\circ}\right)\left(x-\cos100^{\circ}\right)\left(x-\cos140^{\circ}\right)\\ =8\left(x^2-x\cos20^{\circ}-x\cos100^{\circ}+\cos20^{\circ}\cos100^{\circ}\right)\left(x-\cos140^{\circ}\right)\\ =8\left(x^3-x^2\cos20^{\circ}-x^2\cos100^{\circ}+x\cos20^{\circ}\cos100^{\circ}-x^2\cos140^{\circ}+x\cos20^{\circ}\cos140^{\circ}+x\cos100^{\circ}\cos140^{\circ}\\-\cos20^{\circ}\cos100^{\circ}\cos140^{\circ}\right)\\ =8x^3-\color{#3D99F6}{8(\cos20^{\circ}+\cos100^{\circ}+\cos140^{\circ})}x^2+\color{#D61F06}{8(\cos20^{\circ}\cos100^{\circ}+\cos20^{\circ}\cos140^{\circ}+\cos100^{\circ}\cos140^{\circ})}x \\-\color{#EC7300}{8\cos20^{\circ}\cos100^{\circ}\cos140^{\circ}}$

From here on, I will only use these tools:

1. Product to sum trigonometric formula : $\color{#20A900}{2\cos A\cos B=\cos(A+B) +\cos(A-B)}$
2. $\color{magenta}{\cos 60^{\circ} = \dfrac{1}{2}}$
3. For angles $90^{\circ} < x < 180^{\circ}$ , $\color{#69047E}{\cos x = -\cos(180^{\circ}-x)}$
4. For angles $180^{\circ} < x < 270^{\circ}$ , $\color{#624F41}{\cos x = -\cos(x-180^{\circ})}$

We now proceed to determine the values of the coefficients:

$\color{#3D99F6}{8(\cos20^{\circ}+\cos100^{\circ}+\cos140^{\circ})}\\ =8(\color{#20A900}{\cos(60^{\circ}-40^{\circ})+\cos(60^{\circ}+40^{\circ})}+\cos140^{\circ})\\ =8(\color{#20A900}{2\cos60^{\circ}\cos40^{\circ}}+\cos140^{\circ})\\ =8(2\left(\color{magenta}{\dfrac{1}{2}}\right)\cos40^{\circ} +(\color{#69047E}{-\cos(180^{\circ}-140^{\circ})}))\\ =8(\cos40^{\circ}-\cos40^{\circ})\\ =8(0)\\ =\color{#3D99F6}{0}$

$\color{#D61F06}{8(\cos20^{\circ}\cos100^{\circ}+\cos20^{\circ}\cos140^{\circ}+\cos100^{\circ}\cos140^{\circ})}\\ =4(\color{#20A900}{2\cos20^{\circ}\cos100^{\circ}}+\color{#20A900}{2\cos20^{\circ}\cos140^{\circ}}+\color{#20A900}{2\cos100^{\circ}\cos140^{\circ}})\\ =4(\color{#20A900}{\cos80^{\circ}+\cos120^{\circ}}+\color{#20A900}{\cos120^{\circ}+\cos160^{\circ}}+\color{#20A900}{\cos40^{\circ}+\cos240^{\circ}})\\ =4(\color{#20A900}{\cos(60^{\circ}+20^{\circ})}\color{#69047E}{-\cos60^{\circ}-\cos60^{\circ}}+\cos160^{\circ}+\color{#20A900}{\cos(60^{\circ}-20^{\circ})}\color{#624F41}{-\cos60^{\circ}})\\ =4(\color{#20A900}{2\cos60^{\circ}\cos20^{\circ}}+\cos160^{\circ}-3\color{magenta}{\cos60^{\circ}})\\ =4(2\left(\color{magenta}{\dfrac{1}{2}}\right)\cos20^{\circ} \color{#69047E}{-\cos20^{\circ}}-3\left(\color{magenta}{\dfrac{1}{2}}\right))\\ =4(\cos20^{\circ}-\cos20^{\circ}-\dfrac{3}{2})\\ =4\left(-\dfrac{3}{2}\right)\\ =\color{#D61F06}{-6}$

$\color{#EC7300}{8\cos20^{\circ}\cos100^{\circ}\cos140^{\circ}}\\ =4(\color{#20A900}{2\cos20^{\circ}\cos100^{\circ}})\cos140^{\circ}\\ =4\cos140^{\circ}(\color{#20A900}{\cos80^{\circ}+\cos120^{\circ}})\\ =2(\color{#20A900}{2\cos80^{\circ}\cos140^{\circ}}+2\cos140^{\circ}(\color{#69047E}{-\cos60^{\circ}}))\\ =2(\color{#20A900}{\cos60^{\circ}+\cos220^{\circ}}-2\cos140^{\circ}\left(\color{magenta}{\dfrac{1}{2}}\right))\\ =2(\color{magenta}{\dfrac{1}{2}}\color{#624F41}{-\cos40^{\circ}}-(\color{#69047E}{-\cos40^{\circ}}))\\ =1+2(-\cos40^{\circ}+\cos40^{\circ})\\ =\color{#EC7300}{1}$

Substitute the found values into $f(x)$ :

$f(x) = 8x^3-\color{#3D99F6}{8(\cos20^{\circ}+\cos100^{\circ}+\cos140^{\circ})}x^2+\color{#D61F06}{8(\cos20^{\circ}\cos100^{\circ}+\cos20^{\circ}\cos140^{\circ}+\cos100^{\circ}\cos140^{\circ})}x\\ -\color{#EC7300}{8\cos20^{\circ}\cos100^{\circ}\cos140^{\circ}}\\ =8x^3-\color{#3D99F6}{0}x^2+(\color{#D61F06}{-6})x -\color{#EC7300}{1}\\ =8x^3-6x-1$

$f(0)=8(0)^3-6(0)-1=-1\\ f(1)=8(1)^3-6(1)-1=1\\ f(10)=8(10)^3-6(10)-1=7939\\ f(0)+f(1)+f(10)=-1+1+7939=\boxed{7939}$

The eighth degree chebyshev polynomial has factors $8x^3-6x-1$ and $8x^3+6x-1$ . The roots of $8x^3-6x-1$ are the given. Then we sum the values to get $7939$ .