$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, 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.

If you start with the Sum-to-Product Identities, you can rewrite the given equations as
$$2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=-\frac{1}{3} \quad \text{and} \quad -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=\frac{1}{2}.$$
Dividing the second equation by the first one, we'll cancel out $\sin\left(\frac{x-y}{2}\right)$, and as a result we'll find the value of $\tan\left(\frac{x+y}{2}\right)$. Then to finish it up, use one of the half-angle substitutions:
$$\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}.$$
