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Let $\theta=2\pi/7$, show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$

I have found that $\cos(4\theta) = \cos(8\pi/7) = \cos(6\pi/7) = \cos(\theta)$ and, similarly, $\cos(5\theta) = \cos(2\theta)$ and $\cos(6\theta) = \cos(\theta)$.Thus the expression $$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$$ can be written as

$$1+2\cos(\theta)+2\cos(2\theta)+2\cos(3\theta)= 0$$

Where do I get that $\cos(\theta)$ satisfies the polynomial $8x^3+4x^2-4x-1$.

But at the moment I don't see how it can help me with what I need. Is it the correct way? If not, could you give me a hint of where to go?

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If you can use the property that $\cos(x) = Re(e^{i x})$, where $Re(z)$ is the real part of $z$, then:

\begin{align*} \sum_{k=0}^6 \cos(k\theta) &= Re\left(\sum_{k=0}^6 e^{ik\theta}\right) \\ &= Re\left(\frac{1-e^{i7\theta}}{1-e^{i\theta}}\right) \\ &= Re\left(\frac{1-e^{i2\pi}}{1-e^{i\frac{2\pi}{7}}}\right) \\ &= 0 \end{align*} as $e^{i2\pi} = 1$.

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