Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$?  Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\begin{align}
\cos x \cos y &=4/15\\
\sin x \sin y &=-1/15
\end{align}
Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have
$$
225 \sin^4 y -210 \sin^2 y +1=0
$$
with its solution 
$\sin^2 y = \frac{7\pm4\sqrt3}{15}$.
Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.
\begin{align}
\sin^2(2y) &= 4\cos^2 y\sin^2 y\\
&= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\
\sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)} 
\end{align}
$\sin 2y$ must be negative.
Edit
Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.
By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is. 

As $90<x,y<180$ and $\cos(x+y)>0$
$$270<x+y<360\implies\sin(x+y)=-\sqrt{1-(1/3)^2}$$
Again, $-90<x-y<90^\circ\implies\sin(x-y)=\pm\sqrt{1-(1/5)^2}$
Finally $$\sin2y=\sin(x+y+(x-y))=?$$
