Originally Posted by

**satis** The equation is

$\displaystyle 8sin^2xcos^2X = 1-cos4x$

I wasn't sure what to do with the right side, so I did all my work on the left. I factored out the 8

$\displaystyle 8(sin^2xcos^2x)$

which is

$\displaystyle 2(2sinxcosx)(2sinxcosx)$

since 2sinxcosx is the double angle identity for sine I subbed that in

$\displaystyle 2(sin2x)(sin2x)$

which gets turned into

$\displaystyle 2sin^2x$

I may have just gone down the wrong path as I'm not sure how that's supposed to turn into 1-cos4x. Any pointers?