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.
 A: In your  calculation of $x+y$ via $\sin(x+y)$, you wrote that $x+y=\frac{\pi}{4}$ is a possibility. It is not, since already $\sin y=\frac{9}{\sqrt{130}}\gt\frac{1}{\sqrt{2}}$, so $y\gt \frac{\pi}{4}$.
Remark: In this case, computing $\cos(x+y)$ is the better strategy, since the cosine here unambiguously identifies $x+y$. Computing $\sin(x+y)$ also works, but requires an extra step to disambiguate. 
A: Simply: Solving for an angle using its sine value is ambiguous. (This is one reason that the Law of Cosines is preferred over the Law of Sines for determining angles from sides.) 

Here's an analogous situation:
$$\begin{align}x^2 + 2 x = 35 \quad(1)\\
x^2 - 2 x = 15 \quad(2)
\end{align}$$
Adding $(1)+(2)$, we see that $2 x^2 = 50$, so that $x = \pm 5$. On the other hand, $(1)-(2)$ gives $4x = 20$, so that $x = 5$. Two valid approaches seem to give different solution sets. But note: in the first case, one of the proposed values of $x$ is extraneous. You can check this by substitution: $-5$ is not a solution to $(1)$ or $(2)$.

In your problem, you can likewise determine that one of the values from the sine-based solution is extraneous.
We're given that $\cos 2x$ is negative, while $x$ itself is strictly positive and acute; therefore, $2x$ must be strictly larger than a right angle, which implies that $x > \pi/4$. Since $y$ is also assumed positive, it cannot be that $x+y = \pi/4$. Thus, $3\pi/4$ is the only valid solution.

The lesson here is that some approaches to a problem introduce extraneous solutions. You should always double-check that proposed solutions actually work in the original equations.
A: 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.
A: To know which quadrant $x+y$ is in, you need to know the signs of both $\sin(x+y)$ and $\cos(x+y)$ In your case, $\sin(x+y) \gt 0$ and $\cos(x+y) \lt 0$. This puts $x+y$ in the second quadrant. So the solution is the one angle $x+y = \dfrac{3 \pi}4$.
