Nice Exercise asked in many facebook math groups

Exercise

Show that, $\sin 6\sin 54\sin 66=\frac{\sqrt{5}-1}{16}$


Solution:

We know that, $\sin a\sin b=\frac{1}{2}\left( \cos (a-b)-\cos (a+b\right) )$ ,then

$\sin 6\sin 66=\frac{1}{2}\left( \cos (6-66\right) -\cos (6+66))$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \cos \left( -60\right)
-\cos \left( 72\right) \right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \cos 60-\cos 72\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( \frac{1}{2}-\cos 72\right)
=\frac{1}{4}-\frac{1}{2}\cos 72$

hence, $\sin 6\sin 54\sin 66=\frac{1}{4}\sin 54-\frac{1}{2}\sin 54\cos 72$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =%
\frac{1}{4}\sin \left( 90-36\right) -\frac{1}{2}\sin \left( 90-36\right)
\cos \left( 90-18\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{4}\cos 36-\frac{1}{2}\cos 36\sin 18$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{1}{2}-\sin 18\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{1}{2}-\frac{\sqrt{5}-1}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1%
}{2}\cos 36\left( \frac{2-\sqrt{5}+1}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{%
2}\cos 36\left( \frac{3-\sqrt{5}}{4}\right) $

but $\sin 54=\sin \left( 3\times 18\right) ~$\ and $\sin \left( 5\times
18\right) =\sin 90=1$

however, $\sin 36=\sin \left( 90-54\right) =\cos 54$

Let $\theta =18$ then $\ \sin 2\theta =\cos 3\theta $

but $\cos 3\theta =4\cos ^{3}\theta -3\cos \theta $

hence, $\sin 2\theta =4\cos ^{3}\theta -3\cos \theta \Leftrightarrow 2\cos
\theta \sin \theta =4\cos ^{3}\theta -3\cos \theta $

$\Rightarrow \cos \theta \left( 2\sin \theta -4\cos ^{2}\theta +3\right) =0$

$\Rightarrow \cos \theta =0$ \ or \ $2\sin \theta -4\left( 1-\sin ^{2}\theta
\right) +3=0$

but $\theta \neq 0,2\pi $ hence, $2\sin \theta -4+4\sin ^{2}\theta +3=0$

$\Leftrightarrow \left( 2\sin \theta \right) ^{2}+2\sin \theta
-1=0\Leftrightarrow \left( 2\sin \theta \right) ^{2}+\frac{1}{2}\left(
2\right) \left( 1\right) \left( 2\sin \theta \right) +\frac{1}{4}-\frac{1}{4}%
-1=0$

$\Leftrightarrow \left( 2\sin \theta +\frac{1}{2}\right) ^{2}=\frac{1}{4}+1=%
\frac{5}{4}\Leftrightarrow 2\sin \theta =\frac{-1\pm \sqrt{5}}{2}%
\Leftrightarrow \sin \theta =\frac{-1\pm \sqrt{5}}{4}....(1)$

but $\sin 18\in \left[ 0,90\right] ~~$hence $\sin 18=\frac{\sqrt{5}-1}{4}>0$

However, $\cos 36=\cos \left( 2\theta \right) =1-2\sin ^{2}\theta =1-2\left(
\frac{\sqrt{5}-1}{2}\right) ^{2}=\frac{\sqrt{5}+1}{4}$

Therefore, $A=\sin 6\sin 54\sin 66=\frac{1}{2}\cos 36\left( \frac{3-\sqrt{5}%
}{4}\right) $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\sqrt{5}+1}{8}\times \frac{%
3-\sqrt{5}}{4}=\frac{2\sqrt{5}-2}{32}=\frac{\sqrt{5}-1}{16}$