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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)ąsin(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
