$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?

Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\sin(x) - \sin(y))(\cos(x) -  \cos(y)) = -\frac{1}{6} $$
$$ \sin(x)\cos(x) + \sin(y) \cos(y) - \sin(x+y) = \frac{1}{6}$$
thus
$$ \sin(x+y) =\sin(x)\cos(x) + \sin(y) \cos(y) - \frac{1}{6}  $$
Next if I do a sum and organize and squaring we get:
$$ (\sin(x) + \cos(x))^{2}  + (\sin(y) + \cos(y))^{2} - 2 (\sin(x) + \cos(x)) (\sin(y) + \cos(y)) = \frac{1}{36} $$
$$ 2 + 2 ( \sin(x) \cos(x) + \sin(y) \cos(y))  - 2 \left( \sin(x+y) + \cos(x-y) \right) = \frac{1}{36} $$
I have no idea after this.

Another method is I suspect that we have to solve for $\sin(x), \cos(x), \sin(y), \cos(y)$ instead of directly finding $\sin(x+y)$ algebraically.

We have 
$$\cos \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{6}$$
$$\sin \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{4}$$
Thus
$\tan \frac{x+y}{2}=\frac{3}{2}$.
And hence
$$\sin(x+y)=\frac{2\tan\frac{x+y}{2}}{1+\tan^2\frac{x+y}{2}}=\frac{12}{13}$$
