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