<s>[INST] How do you prove: $\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})} = 8\sin(\frac{\pi}{18})$ I have to prove:
$\displaystyle \tag*{} \alpha={\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})}}= 8\sin(\frac{\pi}{18})$
I tried to make a common denominator of $\alpha$ and use a few identities to arrive at:
$\displaystyle \tag*{} \alpha = \dfrac{\sin(\frac{6\pi}{9})}{\sin(\frac{2\pi}{9})\sin^2(\frac{4\pi}
{9})}$
I am not sure how to proceed from this. I even arrived at other 2 forms of $\alpha$, but they make even more complicated. Any hints would be greatly appreciated. Thanks.
 [/INST] Let $c_n:=\cos\frac{n\pi}{9},\,s_n:=\sin\frac{n\pi}{9}$ so we want to prove$$\alpha:=\frac{s_6}{s_2s_4^2}=8c_4.$$Since $s_{9-n}=s_n$ and $s_3=\frac{\sqrt{3}}{2}$,$$\frac{\alpha}{8c_4}=\frac{\sqrt{3}}{8s_1s_2s_4}.$$Now we just need to prove$$s_1s_2s_4=\frac{\sqrt{3}}{8},$$which is a duplicate.
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