In any triangle, $b^2\sin(2\gamma)+c^2\sin(2\beta)=2ac\sin(\beta)$
Solution 1:
Hint (use double-angle formula to expand the left side first):
$1$. By law of sines, we have $b \sin(\gamma)=c \sin(\beta)$;
$2$. Construct the height of edge $a$ and prove $a=b \cos(\gamma)+c \cos(\beta)$.