This is one tough identity! It requires very unusual substitutions,
additions and subtractions of the same quantities.

2sin(x+y)sin(x-y) = cos(2y)-cos(2x)

                  = cos(y+y) - cos(x+x)
             
                  = cos(y+y+x-x) - cos(x+x+y-y)

                  = cos[(x+y)-(x-y)] - cos[(x+y)+(x-y)]

     Use the identities cos(A∓B)=cos(A)cos(B)±sin(A)sin(B)
     with A=(x+y), B=(x-y)

= [cos(x+y)cos(x-y)+sin(x+y)sin(x-y)] - [cos(x+y)cos(x-y)-sin(x+y)sin(x-y)]    

= cos(x+y)cos(x-y) + sin(x+y)sin(x-y) - cos(x+y)cos(x-y) + sin(x+y)sin(x-y)

= cos(x+y)cos(x-y) + sin(x+y)sin(x-y) - cos(x+y)cos(x-y) + sin(x+y)sin(x-y)

                 = 2sin(x+y)(sin(x-y)

Edwin