<s>[INST] Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:

Given  $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.

The first couple of steps are the same: finding  $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}} 
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65}   \\
\sin x &=\frac {8} {\sqrt{65}} 
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2} 
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$  OR  $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant. 
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$ 
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $   OR   $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
 [/INST] Andre already gave the answer, but let me explain the "generalities". 
The main problem is the following:

After you reached the point $\sin(x+y) = \sqrt{2}/2$, you concluded that $x+y$ can take both values $\pi/4$ or $3\pi/4$. 

You interpreted this to mean 

BOTH $x+y = \pi/4$ and $x+y = 3\pi/4$ are solutions. (That is, the set of solutions is the set $\{\pi/4, 3\pi/4\}$.)

This interpretation is false! The correct interpretation is that

We have ruled out all numbers other than $\pi/4$ and $3\pi/4$ from being possible solutions. (Or, the set of solutions is a subset of the set $\{\pi/4, 3\pi/4\}$.)

There may be other constraints that rule out one (or both) of those values. In your case, knowing that $\cos x \approx 1/8$ should tell you already that $x$ is bigger than $\pi/4$. 

Another way to look at this clearly is to look at the directions of implication. From the values of $\sin x$, $\sin y$ etc you can derive the value of $\sin(x+y)$. But just knowing the value of $\sin(x+y)$ you cannot derive the value of $\sin x$, $\sin y$, etc. So the implication going from the "necessary information" step to the "computing $\sin(x+y)$ and $\cos(x+y)$" step is one that loses information. 
So that solutions to $\sin(x+y) = \sqrt{2}/2$ are not necessarily solutions to $\cos(x) = 1/\sqrt{65}$ etc.
</s>