(sina-sin3a/sin2a)*(cosa-cos3a/cos2a) надо

((sinα-sin3α)/sin2α)·((cosα-cos3α)/cos2α=
=[2·sin((α-3α)/2)·cos((α+3α)/2)/sin2α]·[-2sin(-α)·sin2α/cos2α]=
=[-2·sinα·cos2α/sin2α]·[2sinα·sin2α/cos2α]=
=(-4)·sin²α·(1/2·sin4α)/(1/2·sin4α)=
=-4sin²α;