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$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$

I tried showing the equation, but my attempts did not get the result. I already showed that

$$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})=\frac18$$ $$\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=-\frac12$$

Using in particular the trigonometric formulas:

$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ , $\sin(\pi+\alpha)=-\sin(\alpha)$ , $\sin(\pi-\alpha)=\sin(\alpha)$ , $\cos(\pi-\alpha)=-\cos(\alpha)$ , $\sin(-\alpha)=-\sin(\alpha)$ , and $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$

Any hint would be helpful. Thanks in advance.

Quanto
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Dada
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3 Answers3

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$$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$ are the roots of $$ x^3 + x^2 - 2x - 1 $$ so $$ \cos \left( \frac{2 \pi}{7} \right) \; , \; \; \cos \left( \frac{4 \pi}{7} \right) \; , \; \; \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$ are the roots $$ 8 x^3 + 4 x^2 - 4x - 1 $$ or $$ x^3 + \frac{1}{2}x^2 - \frac{1}{2} x - \frac{1}{8} $$ The sum of products of two roots ate a time is $-\frac{1}{2}$

The original claim is from a method due to Gauss, this time we take $$ \omega = e^{2 \pi i / 7} , $$ next take $$ \eta = \omega + \frac{1}{\omega}, $$ then look for a cubic (in $\eta$) that vanishes because of the relation $$ \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$$ Here we find $$ \eta^3 + \eta^2 - 2\eta - 1 = \frac{ \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1}{\omega^3} $$

Will Jagy
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    +1 That's my favorite way of dealing with such trig identities, though it's likely that OP is unfamiliar with this. – Calvin Lin Dec 09 '19 at 18:07
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HINT: $$cos\theta_1cos\theta_2=\frac{cos(\theta_1 - \theta_2) + cos(\theta_1 + \theta_2)}{2}$$

Ty Jensen
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Use the identity $\cos(a-b)+\cos(a+b) = 2\cos a\cos b$, $$I=\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} $$ $$=\frac12\left(\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}\right) +\frac12\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}\right) +\frac12\left(\cos\frac{2\pi}{7}+\cos\frac{10\pi}{7}\right) $$

Recognize $\cos\frac{8\pi}{7} = \cos\frac{6\pi}{7}$ and $\cos\frac{10\pi}{7} = \cos\frac{4\pi}{7}$ to simplify $I$, $$I=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12$$ where the result you already obtained for the sum is used.

Quanto
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