Question 253255
Recall that *[Tex \LARGE sin(x+y)=sin(x)cos(y)+cos(x)sin(y)]. So this means that *[Tex \LARGE sin(x-y)=sin(x)cos(y)-cos(x)sin(y)]




*[Tex \LARGE sin(x+y)sin(x-y)=cos^2(y)-cos^2(x)] ... Start with the given equation.



*[Tex \LARGE \left(sin(x)cos(y)+cos(x)sin(y)\right)\left(sin(x)cos(y)-cos(x)sin(y)\right)=cos^2(y)-cos^2(x)] ... Use the identities given above to expand the left side



*[Tex \LARGE \left(sin(x)cos(y)\right)^2-\left(cos(x)sin(y)\right)^2=cos^2(y)-cos^2(x)] ... FOIL



*[Tex \LARGE sin^2(x)cos^2(y)-cos^2(x)sin^2(y)=cos^2(y)-cos^2(x)] ... Distribute the exponent.



*[Tex \LARGE \left(1-cos^2(x)\right)cos^2(y)-cos^2(x)\left(1-cos^2(y)\right)=cos^2(y)-cos^2(x)] ... Use the identity *[Tex \LARGE sin^2(x)=1-cos^2(x)]



*[Tex \LARGE cos^2(y)-cos^2(x)cos^2(y)-cos^2(x)+cos^2(x)cos^2(y)=cos^2(y)-cos^2(x)] ... Distribute



*[Tex \LARGE cos^2(y)-cos^2(x)=cos^2(y)-cos^2(x)] ... Combine like terms.



Since this equation is true for all values of x and y, this shows that the original equation is true for all values of x and y.



So this verifies that *[Tex \LARGE sin(x+y)sin(x-y)=cos^2(y)-cos^2(x)] is an identity.