Simplify $4^3\sin^4(20^\circ)\sin^2(70^\circ)-4\sqrt3\sin^3(20^\circ)\sin(70^\circ)+3$ I was trying to solve a question where the two sides of a triangle were $$\frac{a\sin(20^\circ)}{\sin(70^\circ)}$$and $$\frac{a\sin(60^\circ)\sin(30^\circ)}{\sin(70^\circ)\sin(40^\circ)}$$ and the ange between them was $70^\circ$ I used the law of cosines to try find the third side which I call $c$ and I soon found that $$\frac{a^2[4^3\sin^4(20^\circ)\sin^2(70^\circ)-4\sqrt3\sin^3(20^\circ)\sin(70^\circ)+3]}{4^3\sin^4(70^\circ)\sin^2(20^\circ)}=c^2$$ But after this I’m not able to simplify further, please help, there still is a possibility that I might’ve done something wrong that I reached this step, please assist.
 A: We have the following theorem:
$$
\prod_{0<k<n}2\sin\frac{k\pi}n=n.
$$
Particularly:
$$\begin{align}
\prod_{0<k<9}{\sin\frac{k\pi}9}=[2^4\sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)]^2=9\\
\implies \sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)=\frac{3}{16}.\tag1
\end{align}
$$
Let us apply this to your triangle:
$$AB=\frac{a\sin(20^\circ)}{\sin(70^\circ)},\quad AC=\frac{a\sin(60^\circ)\sin(30^\circ)}{\sin(70^\circ)\sin(40^\circ)},\\ \alpha=\measuredangle CAB=70^\circ,\quad\beta=\measuredangle ABC,\quad \quad\gamma=\measuredangle BCA.$$
Then we have by law of sines:
$$
\frac{\sin\beta}{\sin\gamma}=\frac{AC}{AB}=\frac{\sin(60^\circ)\sin(30^\circ)}{\sin(20^\circ)\sin(40^\circ)}\stackrel{(1)}
=\frac{16\sin(60^\circ)\sin(30^\circ)\sin(60^\circ)\sin(80^\circ)}3
=\frac{\sin(80^\circ)}{\sin(30^\circ)}.
$$
In view of $\beta+\gamma=110^\circ$ one concludes 
$$\beta=80^\circ,\quad \gamma=30^\circ.$$
Finally:
$$
\frac{BC}{AB}=\frac{\sin\alpha}{\sin\gamma}\implies BC=\frac{a\sin(20^\circ)}{\sin(70^\circ)}\frac{\sin(70^\circ)}{\sin(30^\circ)}=2a\sin(20^\circ).
$$
