Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:

Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$

Here are the two solutions that I found:

I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?

Using Complex Addition
$$
e^{ia}+e^{i(\pi-b)}=\sqrt{\frac12}+i\sqrt{\frac32}=\color{#C00}{\sqrt2}e^{i\color{#090}{\pi/3}}
$$


*

*$b+a=\frac\pi2$ since $\left|e^{ia}+e^{i(\pi-b)}\right|^2=\color{#C00}{2}\implies\overbrace{\color{#C00}{2}=2+2\cos((\pi-b)-a)}^{\text{Law of Cosines}}$

*$b-a=\frac\pi3$ since $\frac{a+(\pi-b)}2=\color{#090}{\frac\pi3}$
Therefore,
$$
a=\frac\pi{12}\text{ and }b=\frac{5\pi}{12}
$$
