<s>[INST] 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.
 [/INST] Formulae for $\sin4x,\,\sin5x$ provide another solution. Let $S:=\sin^220^\circ\ne\sin^260^\circ=\tfrac34$ so$$\begin{align}0&=\frac{\sin^2100^\circ-\sin^280^\circ}{S}\\&=(5-20S+16S^2)^2-16(1-S)(1-2S)^2\\&=256S^4-576S^3+432S^2-120S+9\\&=(4S-3)(64S^3-96S^2+36S-3)\\\implies64S^3-96S^2+36S-3&=0\\\implies\frac{4}{S}-\frac{1}{S(1-S)}+64S&=\frac{3-4S+64S^2-64S^3}{S(1-S)}\\&=32.\end{align}$$
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