given the graph 
Use the cross product $U\times V$ to prove the identity $$\sin(a-b) = \sin(a)\cos(b) - \cos(a)\sin(b).$$
I tried to solve but ended with the solution $\sin(a-b) = \cos(a)\sin(b) - \sin(a)\cos(b)$ with the work i did provided below
$$\begin{split} U\times V &= |U||V| \sin(a-b) e \\ U\times V &= (|U||V|) \sin(a-b) \frac{U \times V}{|U\times V|}\\ \frac{|U\times V|}{|U||V|} &= \sin(a - b)\\ \frac{U_x V_y - U_yV_x}{|U||V|}&= \sin(a-b)\\ \frac{\cos(a)U \sin(b)V - \sin(a)U \cos(b)V}{|U||V|}&= \sin(a-b) \\ \cos(a) \sin(b) - \sin(a) \cos(b)) &= \sin(a-b). \end{split}$$