Official solution
sin 10\^{\circ} sin30\^{\circ} sin50\^{\circ} sin70\^{\circ}
= sin30\^{\circ} sin50\^{\circ} sin 10\^{\circ} sin70\^{\circ}
= \[{1 \over 2}\] [ sin50\^{\circ} sin 10\^{\circ} sin70\^{\circ} ]
= \[{1 \over 2}\] [ sin(60\^{\circ} - 10\^{\circ}) sin 10\^{\circ} sin(60\^{\circ} + 10\^{\circ}) ]
= \[{1 \over 2}\] [ \[{1 \over 4}\sin \]3(10\^{\circ}) ]
= \[{1 \over 2}\] [ \[{1 \over 4} \times {1 \over 2}\]]
= \[{1 \over {16}}\]
Note :
\[\sin \left( {60^\circ - A} \right)\sin A\sin \left( {60^\circ - A} \right) = {1 \over 4}\sin 3A\]