Recall that
\(\sin(x+y)=\sin x\cos y+\cos x\sin y\)
\(\sin(x-y)=\sin x\cos y-\cos x\sin y\)
Adding these together, you have
\(\sin(x+y)+\sin(x-y)=2\sin x\cos y\)
\(\dfrac12\sin(x+y)+\dfrac12\sin(x-y)=\sin x\cos y\)
Replace \(x\) with \(2x\) and \(y\) with \(3x\). You end up with
\(\dfrac12\sin(2x+3x)+\dfrac12\sin(2x-3x)=\sin2x\cos3x\)
\(\sin2x\cos3x=\dfrac12\sin5x+\dfrac12\sin(-x)=\dfrac12\sin5x-\dfrac12\sin x\)
and so \(A=\dfrac12\) and \(B=-\dfrac12\).
