If $\sin \mathrm{A}+\sin \mathrm{B}=x$ and $\cos \mathrm{A}+\cos \mathrm{B}=\mathrm{y}$, then $\sin (\mathrm{A}+\mathrm{B})=$

Given: $\sin A + \sin B = x$ and $\cos A + \cos B = y$.

Apply sum-to-product identities:

$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = x$

$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = y$

Dividing these equations (assuming $\cos\left(\frac{A-B}{2}\right) eq 0$):

$\tan\left(\frac{A+B}{2}\right) = \frac{x}{y}$

Using the tangent half-angle identity for sine:

$\sin(A+B) = \frac{2 \tan\left(\frac{A+B}{2}\right)}{1 + \tan^2\left(\frac{A+B}{2}\right)} = \frac{2(x/y)}{1 + (x/y)^2}$

Simplify:

$\sin(A+B) = \frac{2x/y}{(x^2 + y^2)/y^2} = \frac{2x}{y} \cdot \frac{y^2}{x^2 + y^2} = \frac{2xy}{x^2 + y^2}$

Result: $\frac{2xy}{x^2+y^2}$