I think I got this right, but may be missing some steps or may be wrong entirely. The idea is to use the double angle identity to simplify

$\displaystyle sin4x$

$\displaystyle = 2sin2xcos2x$

$\displaystyle = 2(2sinxcosx)(1-2sin^2x)$

$\displaystyle = 4sinxcosx(1-2sin^2x)$

$\displaystyle = 4sinxcosx-8sin^3xcosx$

Is this right?