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.
 A: 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.
A: Good question. For me, I am more customized to write in degrees. Therefore, we can have
$${\frac{1}{\sin ^2 (40^\circ)} - \frac{1}{\sin ^2(80^\circ)}}= 8\sin(10^\circ)$$
Or we can write it in a different way
$${\frac{\sin ^2(80^\circ)}{\sin ^2 (40^\circ)}} - 1= 8\sin(10^\circ)\sin ^2(80^\circ)$$
Notice that $$\sin 20^\circ=2\sin10^\circ\cos10^\circ=2\sin10^\circ\sin 80^\circ$$
Also we have $$\sin 80^\circ=2\sin 40^\circ\cos40^\circ$$
So we ned to show
$$4\cos^2 40^\circ -1=4\sin 20^\circ\sin 80^\circ$$
Notice that $1=2\sin30^{\circ}$. so we have
\begin{align}&4\sin 20^\circ\sin 80^\circ\\=&4\sin (50+30)^\circ\sin (50-30)^\circ\\=&4(\sin 50^\circ\cos30^\circ+\sin 30^\circ\cos50^\circ)(\sin 50^\circ\cos30^\circ-\sin 30^\circ\cos50^\circ)\\=&4\sin^2 50^\circ\cos^230^\circ-\sin^2 30^\circ\cos^250^\circ)\\=&4(\sin^2 50^\circ-\sin^230^\circ)\\=&4\sin^2 50^\circ-4\sin^230^\circ=4\cos^2 40^\circ -1
\end{align}
