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.
$\endgroup$ 14 Answers
$\begingroup$Denote $Q = \sin 20 \sin 40 \sin 60 \sin 80$. Observe first that
\begin{align} \sin 60 & = \sin (20+40) \\ & = \sin 20 \cos 40 + \cos 20 \sin 40 \\ & = \sin 20 (2\cos^2 20 - 1) + \cos 20 (2 \sin 20 \cos 20) \\ & = (4\cos^2 20 - 1) \sin 20 \end{align}
We can now write
\begin{align} Q & = \sin 20 \sin (60-20) \sin 60 \sin (60+20) \\ & = \sin 20 (\sin 60 \cos 20 - \cos 60 \sin 20) \sin 60 (\sin 60 \cos 20 + \cos 60 \sin 20) \\ & = \sin 20 \sin 60 (\sin^2 60 \cos^2 20 - \cos^2 60 \sin^2 20) \\ & = \sin 20 \sin 60 \frac{3\cos^2 20 - \sin^2 20}{4} \\ & = \sin 20 \sin 60 \frac{4\cos^2 20 - 1}{4} \\ & = \sin 60 \frac{(4\cos^2 20 - 1)\sin 20}{4} \\ & = \sin 60 \frac{\sin 60}{4} \\ & = \frac{\sin^2 60}{4} = \frac{3}{16} \end{align}
$\endgroup$ $\begingroup$HINT:
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin x[\sin^2120^\circ-\sin^2x]=\dfrac{3\sin x-4\sin^3x}4$$
$$\implies4\sin(60^\circ-x)\cdot\sin x\sin(60^\circ+x)=\sin3x$$
Set $x=20^\circ$
$\endgroup$ 2 $\begingroup$Hint:
Use the so-called Morrie's law :
$\cos(20°)\cos(40°)\cos(80°)=1/8$
and
$\sin(20°)\sin(40°)\sin(80°)=\dfrac{\sqrt{3}}{8}$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 2
