Question 253255
We have to prove:
(i) sin(x+y)*sin(x-y))=(cos^2y-cos^2x
I will simplify the left into the right.
identity: sin (x+y) = sinxcosy + sinycosx
identity: sin(x-y) = sinxcosy - sinycosx 
from (i) we get
(ii) {{{(sinxcosy + sinycosx)*(sinxcosy - sinycosx)}}}
multiply and we get
(iii) sin^2x*cos^2y - sin^2y*cos^2x
identity: sin^2x = 1 - cos^2x
identity: sin^2y = 1 - cos^2y
(iii) becomes
(iv) (1-cos^2x)(cos^2y) - (1-cos^2y)(cos^2x)
simplifying (iv) we get
(v) cos^2y - cos^2x
QED