I tried a few things. In particular I noticed one fact that might be of significance, that is the sum of the angles $2\pi/7$, $4\pi/7$ and $8\pi/7$ is precisely $=2\pi$. And tried to format $2\sin A\cos A$. However, I was not able to make much progress. Can someone please help me with this?
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2Related to Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$. – peterwhy Nov 03 '23 at 04:52
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We try to find the (or a) cubic polynomial with roots $b_1=\cos(2\pi/7)$, $b_2=\cos(4\pi/7)$ and $b_3=\cos(8\pi/7)$. Note, if we consider $0=1+z+z^2+z^3+z^4+z^5+z^6$, then this polynomial has roots $e^{2i\pi/7}$, $e^{4i\pi/7}$, $e^{6i\pi/7}$, $e^{8i\pi/7}$, $e^{10i\pi/7}$, $e^{12i\pi/7}$. Note $e^{frac{10i \pi}{7}}=e^{frac{-4i \pi}{7}}$ and $e^{\frac{12i \pi}{7}}=e^{\frac{-2i \pi}{7}}$ & so on.
Let $c_1=\cos(\frac{2\pi}{7})+i\sin \frac{2\pi}{7}$ and similarly for $c_2$ and $c_3$. Hence the roots of this polynomial are $\{c_1, c_1, c_2, c_2, c_3, c_3 \}$. Now, as the sum of roots of polynomial of degree $n$ is $-\frac{k_{n-1}}{k_n}$, where $k_{n-1}$ is the coefficient of $z^{n-1}$ and $k_n$ is the coefficient of $z^n$, we see (considering the polynomial $1+z+z^2+z^3+z^4+z^5+z^6$) that $2(e^{2i\pi/7}+e^{4i\pi/7}+e^{8i\pi/7})=-1$. Comparing the real parts on both sides, $-1=2(\cos(2\pi/7)+\cos(4\pi/7)+\cos(8\pi/7))$. So $-\frac{1}{2}=\cos(2\pi/7)+\cos(4\pi/7)+\cos(8\pi/7)$. And, if the polynomial $P(z)=1+z+z^2+z^3+z^4+z^5+z^6$ is re-written as $z^3+\frac{1}{z^3}+z^2+\frac{1}{z^2}+z+\frac{1}{z}+1$, we see (if $v=z+\frac{1}{z}$), then $P(v)=v^3-3v+v^2-2+v+1$, hence $P(v)=v^3+v^2-2v-1$. Note that $P(v)$ has roots $2\cos(\frac{2\pi}{7})$ and $2\cos(\frac{4\pi}{7})$ and $2\cos(\frac{8\pi}{7})$, hence $4(\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{8\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{8\pi}{7}))=-2$. Cheers.
Surajit131
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Given expression
$$ \cos\left(\frac{8\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)\cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) $$
$$ = -\cos\left(\frac{\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) + -\cos\left(\frac{\pi}{7} \right)\cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) $$
$$ = -\cos\left(\frac{\pi}{7} \right)\left( \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) $$
$$ = 2\cos^2\left(\frac{\pi}{7} \right) \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{2\pi}{7} \right) $$
$$ = \cos\left(\frac{4\pi}{7}\right)\left( 2\cos^2\left(\frac{\pi}{7} \right) + \cos\left(\frac{2\pi}{7} \right) \right) = 2\cos\left(\frac{2\pi}{7} \right)\cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) $$
$$ = \cos\left(\frac{6\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{2\pi}{7} \right) = \frac{\sin\left(\frac{6\pi}{7\times 2} \right) \cos\left(\frac{1}{2}\right)\left(2\frac{2\pi}{7} + (3 - 1)\frac{2\pi}{7} \right)}{\sin\left(\frac{\pi}{7} \right)}$$
where we have used the summation of sines and cosines with angles in in arithmetic progressions formula. Further simplifying, we have
$$ \frac{\sin\left(\frac{3\pi}{7} \right) \cos\left(\frac{4\pi}{7}\right)}{\sin\left(\frac{\pi}{7} \right)} = \frac{-\sin\left(\frac{3\pi}{7} \right) \cos\left(\frac{3\pi}{7}\right)}{\sin\left(\frac{\pi}{7} \right)} = \frac{-\sin\left(\frac{6\pi}{7} \right) }{2\sin\left(\frac{\pi}{7} \right)} = \frac{-\sin\left(\frac{\pi}{7} \right) }{2\sin\left(\frac{\pi}{7} \right)} = \frac{-1}{2}$$
I might have missed a trivial way to simplify the expression, for instance the sum of cosines formula is not necessary to obtain the final result, but the manipulations would be somewhat messy so I decided to use that instead.
