Simplify $\sin (x-y) \cos y + \cos (x-y) \sin y$.
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \sin x\qquad \textbf{(C)}\ \cos x\qquad \textbf{(D)}\ \sin x \cos 2y\qquad \textbf{(E)}\ \cos x\cos 2y$
By the addition formula for $\sin$, this becomes $\sin{((x-y)+y)} = \sin{x}$, so the answer is $\boxed{\textbf{(B)}}$.