<s>[INST] Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.
Any hints are welcomed.
 [/INST] We have:
$\sin20^{\circ} = \sin(30^\circ - 10^\circ)= \frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin40^{\circ} = \sin(30^\circ + 10^\circ)= \frac{1}{2}\cos10^\circ + \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin 60^\circ = \frac{\sqrt{3}}{2}$
$\sin80^\circ =\cos 10^\circ$
Therefore,
$A = \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{\sqrt{3}}{2} \left((\cos10^\circ )^2 - \frac{3}{4} \right)\cos10^\circ$
Moreover :
$ \cos30^\circ = 4(\cos10^\circ )^3 - 3\cos 10^\circ$, from this cubic equation we find out the value of $cos10^\circ$, and then plug it to the expression of $A$ we will get the answer.
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