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)?
 A: 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,
$$
a=\frac\pi{12}\text{ and }b=\frac{5\pi}{12}
$$
A: Your first way is right.
In the second solution $$A+\frac{\pi}{6}=B-\frac{\pi}{6},$$ which is very well or $$A+B=0,$$ which is impossible. 
Another way:
$$2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\sqrt{\frac{3}{2}}$$ and
$$2\sin\frac{A+B}{2}\sin\frac{B-A}{2}=\sqrt{\frac{1}{2}}.$$
Thus,
$$\tan\frac{B-A}{2}=\frac{1}{\sqrt3}$$ and since $A$ and $B$ they are acute angles, we obtain $$B-A=60^{\circ},$$
$$\sin\frac{A+B}{2}=\frac{1}{\sqrt2}$$ and
$$A+B=90^{\circ}.$$
A: $$\sqrt3\cos A-\sin A=\sqrt3\cos B+\sin B$$
$$\implies\cos(A+30^\circ)=\cos(B-30^\circ)$$
$$A+30^\circ=360^\circ n\pm(B-30^\circ) $$
$$-\implies A+B=360^\circ n$$ which is impossible as  $0<A+B<180^\circ$
$+\implies A-B=360^\circ n-60^\circ$
As $-90<A-B<90,n=0$
Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html in  anyone of the given relation
A: My question was more about why my second approach yields an incorrect solution, ignoring the arbitrary assertion that $A$ and $B$ must be acute in the final answer selected from the solution set for $A+B$.   I have answered my own question, however, with the help/tips from everyone who kindly provided comments/answers.
Given the values for $A$ and $B$ that robjohn kindly found using Complex Addition (+1) the $ B-A=\frac{\pi}{3} $ solution is correct using my second strategy where this is the antepenultimate step:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
and then:
$$ B-A=\frac{\pi}{3} $$
And $\frac{\pi}{3}$ is indeed an element of the correct solution set for $B - A$
$ $
The $ A+B = 0 $ $ $ solution that you can get by multiplying the inputs of either of the cosine functions (but not the other) by $-1$ is not correct, however; it does not match any valid solutions in the solution set for $A + B$: $ \frac{\pi}{2}+\pi*k , $ $ $ $k \in \mathbb{z} $.   $ $ Also, as one or two answers pointed out, $A$ and $B$ must be acute so $ A + B $ cannot equal $0$ (although I was less concerned with this than I was with the fact that the answer didn't match any of the valid solutions in the solution set for $A+B$).   I now realize that the reason for this is because multiplying the inputs of the cosine functions in such an equation does not change the validity of the equation (because cosine is even) but it does change the value of the variables in that equation.   
In other words:
If $\pm N = \pm M$ then $\cos N = \cos M$.   However, if $\cos N = \cos M$ then $\pm N$ does not necessarily equal $ \pm M$.   $ $ Another example:
Let $M = \frac{\pi}{4}$ $ $ , $ $ Let $N = \frac{7*\pi}{4}$
$ cosM =cosN $
But $M \ne N$
This seems obvious when put this way but I did not realize or think of it when doing out this solution.
