Apply the rules of sum/difference for trigonometric functions:
[tex] \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y)[/tex]
[tex] \cos(x-y) = \sin(x) \sin(y) + \cos(x) \cos(y)[/tex]
Multiply the two expressions:
[tex] (\sin(x) \cos(y) + \cos(x) \sin(y)) (\sin(x) \sin(y) + \cos(x) \cos(y)) = [/tex]
[tex] \sin(x) \cos(y)\sin(x) \sin(y) + \sin(x) \cos(y)\cos(x) \cos(y) + \cos(x) \sin(y)\sin(x) \sin(y) + \cos(x) \sin(y)\cos(x) \cos(y) = [/tex]
[tex] \sin^2(x) \cos(y)\sin(y) + \sin(x) \cos^2(y)\cos(x)+ \cos(x) \sin^2(y)\sin(x) + \cos^2(x) \sin(y)\cos(y) [/tex]
You can factor [tex]\cos(y)\sin(y) [/tex] from the first and last term, and [tex] \sin(x)\cos(x) [/tex] from the two middle terms: you have
[tex] \cos(y)\sin(y)(\sin^2(x)+\cos^2(x)) + \sin(x) \cos(x)(\cos^2(y)+\sin^2(y)) [/tex]
Since [tex] \sin^2(x)+\cos^2(x)=1 [/tex], the identity is proven.
