Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$ Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn't work.
 A: Per the identity
$\sin3t =3\sin t -4\sin^3t$,
it is straightforward to verify that $\sin20$, $\sin40$, $-\sin80$ are the roots of
$4x^3-3x+\frac{\sqrt3}2=0$, which leads to
$$\sin20\sin40\sin80=\frac{\sqrt3}8$$
Then
\begin{align}
& \frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ\\
=& \frac{64}3\left(4\sin^240\sin^280 -4\sin^220\sin^280+3\sin^220 \right)\\
 =& \frac{64}3\left((1-\cos80)(1+\cos20)-(1-\cos40)(1+\cos20) + \frac32(1-\cos40)\right)\\
 =& \frac{64}3\left(\frac32 - \cos80 -\frac12\cos40 
-\cos20\cos80 +\cos20\cos40\right)\\
 =& 32 + \frac{64}3\left(- \cos80 -\frac12\cos40 
-\frac12(\cos60-\cos80) +\frac12(\cos20+\cos60)\right)\\
 =& 32 + \frac{32}3\left(\cos 20- \cos40 -\cos80
\right) \\
= &32 + \frac{32}3\left(\cos 20- 2\cos60\cos20
\right)\\
=& 32
\end{align}
A: Formulae for $\sin4x,\,\sin5x$ provide another solution. Let $S:=\sin^220^\circ\ne\sin^260^\circ=\tfrac34$ so$$\begin{align}0&=\frac{\sin^2100^\circ-\sin^280^\circ}{S}\\&=(5-20S+16S^2)^2-16(1-S)(1-2S)^2\\&=256S^4-576S^3+432S^2-120S+9\\&=(4S-3)(64S^3-96S^2+36S-3)\\\implies64S^3-96S^2+36S-3&=0\\\implies\frac{4}{S}-\frac{1}{S(1-S)}+64S&=\frac{3-4S+64S^2-64S^3}{S(1-S)}\\&=32.\end{align}$$
