Question 359828
We're going to use the established identities


1) {{{sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}}}


2) {{{sin(A-B)=sin(A)cos(B)-cos(A)sin(B)}}}




{{{sin(x+y) - sin(x-y) = 2cos(x)sin(y)}}} Start with the given equation.



{{{sin(x)cos(y)+cos(x)sin(y) - sin(x-y) = 2cos(x)sin(y)}}} Use the first identity (given above) to expand {{{sin(x+y)}}}



{{{sin(x)cos(y)+cos(x)sin(y) - (sin(x)cos(y)-cos(x)sin(y)) = 2cos(x)sin(y)}}} Use the second identity (given above)  to expand {{{sin(x-y)}}}



{{{sin(x)cos(y)+cos(x)sin(y) -sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)}}} Distribute.



{{{(sin(x)cos(y) - sin(x)cos(y))+(cos(x)sin(y)+cos(x)sin(y))= 2cos(x)sin(y)}}} Group like terms.



{{{0+2cos(x)sin(y)=2cos(x)sin(y)}}} Combine like terms.



{{{2cos(x)sin(y)=2cos(x)sin(y)}}} Simplify



So this verifies the identity.



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