Trigonometric equality sin15°sin24°sin57°=sin39°sin27°sin18°

[anemone]

Prove $\sin 15^\circ \sin 24^\circ \sin 57^\circ= \sin 39^\circ \sin 27^\circ \sin 18^\circ$.
This is an unsolved problem I found @ AOPS.

 

[Opalg]

Prove $\sin 15^\circ \sin 24^\circ \sin 57^\circ= \sin 39^\circ \sin 27^\circ \sin 18^\circ$.

[sp]Using the sum-and-product trig formulas, $$\begin{aligned}\sin 15^\circ \sin 24^\circ \sin 57^\circ &= \sin 15^\circ\cdot\tfrac12\bigl( \cos 33^\circ - \cos 81^\circ\bigr) \\ &= \tfrac14\bigl( \sin 48^\circ - \sin 18^\circ - \sin 96^\circ + \sin 66^\circ\bigr),\qquad(1) \\ \\ \sin 39^\circ \sin 27^\circ \sin 18^\circ &= \sin 39^\circ\cdot\tfrac12\bigl( \cos 9^\circ - \cos 45^\circ\bigr) \\ &= \tfrac14\bigl( \sin 48^\circ + \sin 30^\circ - \sin 84^\circ + \sin 6^\circ\bigr). \qquad(2)\end{aligned}$$

Comparing (1) and (2), they both contain $\sin 48^\circ$. Also, $\sin 96^\circ = \sin 84^\circ$ (because the sine function is symmetric on either side of $90^\circ$). So to show that (1) and (2) are equal, we need to prove that $\sin 66^\circ - \sin 18^\circ - \sin 6^\circ = \sin 30^\circ = \frac12.$ But $$\sin 66^\circ - \sin 6^\circ = 2\cos 36^\circ \sin 30^\circ = \cos 36^\circ = \sin 54^\circ.$$

Therefore it remains to show that $\sin 54^\circ - \sin 18^\circ = \frac12.$ The nicest way to do that is to use the geometry of the pentagon.
[TIKZ]
\coordinate [label=above: $C$] (C) at (90:5cm) ;
\coordinate [label=above right: $D$] (D) at (18:5cm) ;
\coordinate [label=above left: $B$] (B) at (162:5cm) ;
\coordinate [label=below: $A$] (A) at (234:5cm) ;
\coordinate [label=below: $E$] (E) at (306:5cm) ;
\coordinate [label=below: $P$] (P) at (-4.755,-4.045) ;
\coordinate [label=below: $Q$] (Q) at (4.755,-4.045) ;
\draw (A) -- node

{$1$} (B) -- node[above left] {$1$} (C) -- node[above right] {$1$} (D) -- node

{$1$} (E) -- node[below] {$1$} (A) ;
\draw (A) -- (P) -- (B) -- (D) -- (Q) -- (E) ;
\draw (C) -- (0,1.545) ;
\draw (-3.4,-3.75) node {$72^\circ$} ;
\draw (3.45,-3.75) node {$72^\circ$} ;
\draw (-5.1,0.8) node {$18^\circ\to$} ;
\draw (5.1,0.8) node {$\leftarrow18^\circ$} ;
\draw (-0.35,4.3) node {$54^\circ$} ;
\draw (0.4,4.3) node {$54^\circ$} ;
[/TIKZ]
In the diagram, $ABCDE$ is a regular pentagon with sides of length $1$, and $PBDQ$ is a rectangle. It is clear from the given angles that $BD = 2\sin 54^\circ$ and that $PQ = 1 + 2\sin 18^\circ$. Therefore $2\sin 54^\circ = 1 + 2\sin 18^\circ$, from which $\sin 54^\circ - \sin 18^\circ = \frac12.$[/sp]​