Given $ \sin A + \sin B = a$ and $ \cos A + \cos B = b$
where $a$ and $b$ are acute angles
find the value of $\cos(A+B)$
This is my approach
$ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A+B}{2} $
$ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A+B}{2} $
dividing yield $\tan\frac{A+B}{2}=\frac{a}{b}$
$\tan(A+B) = \frac {2\frac{a}{b}}{1-\left(\frac{a}{b}\right)^2} = \frac{2ab}{b^2-a^2}$
and
$\cos(A+B)=\sqrt{\frac{1}{1+\left(\frac{2ab}{b^2-a^2}\right)^2}} = \frac{b^2-a^2}{b^2+a^2}$
I wonder if
$\cos(A+B)= \frac{a^2-b^2}{b^2+a^2}$
was right, if not, why is it?