Determining $\sin(2x)$ 
Given that
$$\sin (y-x)\cos(x+y)=\dfrac 1 2$$
$$\sin (x+y)\cos (x-y) = \dfrac 1 3 $$
Determine $\sin (2x)$.

As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\sin(2x)$ is given by 
$$\sin(2x) = 2\sin\cos$$
Let us try simpiflying the second equation
$$\sin(x+y)-\cos(x-y)=\sin x \cos y+\cos x \sin y-\cos x \cos y-\sin x\sin y=
\cos x(\sin y-\cos y)+\sin x(\cos y-\sin y)=\color{blue}{(\cos x-\sin x)(\sin y-\cos y})$$
However, there seems to be nothing useful. 
 A: Hint
$$
\sin 2x=\sin((x+y)+(x-y))=\sin(x+y)\cos(x-y)+\cos(x+y)\sin(x-y)
$$
We know the value of the first summand. For the second use the fact that sine is odd.
A: Recall that by Product to sum identity
$$2\sin \theta \cos \varphi = {{\sin(\theta + \varphi) + \sin(\theta - \varphi)} }$$
that is
$$\begin{cases}\sin (y-x)\cos(x+y)=\frac12\sin (2y)+\frac12 \sin (-2x)=\dfrac 1 2\\\sin (x+y)\cos (x-y) = \frac12\sin (2x)+\frac12 \sin (2y)=\dfrac 1 3\end{cases} $$
$$\begin{cases}
\sin (2y)-\sin (2x)=1\\
\sin (2x)+\sin (2y)=\dfrac 2 3
\end{cases} $$
and subtracting the first equation from the second we obtain
$$2\sin (2x)=-\frac13 \implies \sin (2x)=-\frac16$$
A: $$ 2x = ( x+y) + (x-y)$$
$$ \sin ( \alpha + \beta ) = \sin (\alpha) \cos (\beta) + \cos (\alpha) \sin (\beta) $$
$$ \sin ( 2x ) = \sin (x+y) \cos (x-y) + \cos (x+y) \sin (x-y)=1/3-1/2 =-1/6$$
A: You were on the right track but the equations contain the product, not the difference.
\begin{align}
\frac12 &= \sin (y-x)\cos(x+y)\\
&=(\sin y\cos x - \sin x \cos y)(\cos x\cos y - \sin x\sin y) \\
&= \sin y\cos y - \cos x \sin x
\end{align}
\begin{align}
\frac13 &= \sin (x+y)\cos(x-y)\\
&=(\sin x\cos y + \sin y \cos x)(\cos x\cos y + \sin x\sin y) \\
&= \sin y\cos y + \cos x \sin x
\end{align}
So subtracting them gives
$$\sin 2x = 2\cos x\sin x = ( \sin y\cos y + \cos x \sin x) - ( \sin y\cos y - \cos x \sin x) = \frac13 - \frac12 = -\frac16$$
