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\[\cos(u+v)\sin(v)+\sin(u+v)\cos(v)\] is there a shortcut
this is of the sin(A+B)
hence the given expression reduces to sin(u+v+v)
=sin(u+2v)
yup, there is : cosA sin B + sinA cos B = sin (A+B) so here it would be sin(u+v+v) = sin (u+2v)
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