<s>[INST] If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ If $A= 2 \times \pi/7$ then how to show,
$$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.
 [/INST] Let us consider $\Phi_7(x)=\frac{x^7-1}{x-1}$. It is a palyndromic polynomial, hence $\frac{\Phi_7(x)}{x^3}$ can be written as a polynomial of $x+\frac{1}{x}$. Given that $\Phi_7(x)$ vanishes at the primitive seventh roots of unity, we get a polynomial that vanishes at $\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}\right\}=\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{8\pi}{7}\right\}=A$:
$$ f(x)=8x^3+4x^2-4x-1 = 8\prod_{\zeta\in A}(x-\zeta) \tag{1}$$
from which:
$$ \frac{f'(x)}{f(x)} = \frac{d}{dx}\log f(x) =\sum_{\zeta\in A}\frac{1}{x-\zeta}\tag{2}$$
and:
$$ \sum_{\zeta\in A}\frac{1}{\zeta} = -\frac{f'(0)}{f(0)} = -\frac{-4}{-1} = \color{red}{-4} \tag{3}$$
as wanted.
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