What is the solution to this trigonometric challenge?

[anemone]

Evaluate $\dfrac{\sin^2 \dfrac{\pi}{7}}{\sin^4 \dfrac{2\pi}{7}}+\dfrac{\sin^2 \dfrac{2\pi}{7}}{\sin^4 \dfrac{3\pi}{7}}+\dfrac{\sin^2 \dfrac{3\pi}{7}}{\sin^4 \dfrac{\pi}{7}}$ without the help of a calculator.

 

[anemone]

Spoiler: Solution of other

Let $s_k=\sin \dfrac{k\pi}{7}$ and $c_k=\cos \dfrac{k\pi}{7}$.

We have

$\dfrac{\sin^2 x}{\sin^4 x}=\dfrac{1}{16\sin^2 x \cos^4 x}=\dfrac{1}{16\sin^2 x \cos^2 x}+\dfrac{1}{16\cos^4 x}=\dfrac{1}{16\sin^2 x}+\dfrac{1}{16\cos^2 x}+\dfrac{1}{16\cos^4 x}$

Therefore we want to find $\displaystyle \sum_{k=1}^3 \dfrac{s_k^2}{s_{2k}^4}=\sum_{k=1}^3 \dfrac{1}{16s_k^2}+\sum_{k=1}^3 \dfrac{1}{16c_k^2}+\sum_{k=1}^3 \dfrac{1}{16c_k^4}$

$\sin 7x=\sin x(64\sin^6 x-112\sin^4 x+56\sin^2 x-7)$

Let the polynomial $64x^6-112x^4+56x^2-7$ has roots $\pm s_1,\,\pm s_2,\,\pm s_3$ so that the polynomial $P_1(x)=64x^3-112x^2+56x-7$ has roots $s_1^2,\,s_2^2,\,s_3^2$ and the polynomial $Q_1(x)=7x^3-56x^2+112x-64$ has roots $s_1^{-2},\,s_2^{-2},\,s_3^{-2}$.

$\displaystyle \sum_{k=1}^3 \dfrac{1}{s_k^2}=s_1^{-2}+s_2^{-2}+s_3^{-2}=\dfrac{56}{7}=8$

And

$P_2(x)=-P_1(1-x)=64x^3-80x^2+24x-1$ has roots $c_1^2,\,c_2^2,\,c_3^2$ and the polynomial $Q_2(x)=x^3-24x^2+80x-64$ has roots $c_1^{-2},\,c_2^{-2},\,c_3^{-2}$.

$\displaystyle \sum_{k=1}^3 \dfrac{1}{c_k^2}=c_1^{-2}+c_2^{-2}+c_3^{-2}=24$

$\displaystyle \sum_{k=1}^3 \dfrac{1}{c_k^4}=c_1^{-4}+c_2^{-4}+c_3^{-4}=24^2-2\times 80=416$

$\therefore \displaystyle \sum_{k=1}^3 \dfrac{s_k^2}{s_{2k}^4}=\dfrac{8}{16}+ \dfrac{24}{16}+\dfrac{416}{16}=28.$