A sufficient condition to ensure $\alpha=\beta$ Let $\alpha$, $\beta$ be acute angles satisfying 
$$
\frac{\sin 2\alpha}{\sin(2\alpha+\beta)}=\frac{\sin2\beta}{\sin(2\beta+\alpha)}             
$$
Show then $\alpha=\beta$.
 A: I got a solution for this problem. Here we have to use the following identities
\begin{eqnarray*}
\sin A\sin B&=&-\frac{1}{2}[\cos(A+B)-\cos(A-B)],\\
\cos A-\cos B&=&-2\sin\frac{A+B}{2}\sin\frac{A-B}{2},\\
\sin 3A&=&3\sin A-4\sin^3 A,\\
\sin 5A&=&16\sin^5 A-20\sin^3 A+5\sin A.
\end{eqnarray*}
We obtain
\begin{eqnarray*}
&&\frac{\sin 2\alpha}{\sin(2\alpha+\beta)}=\frac{\sin 2\beta}{\sin(2\beta+\alpha)}\\
&\Leftrightarrow&\sin 2\alpha\sin(2\beta+\alpha)=\sin 2\beta\sin(2\alpha+\beta) \\
&\Leftrightarrow&-\frac{1}{2}[\cos(3\alpha+2\beta)-\cos(\alpha-2\beta)]=-\frac{1}{2}[\cos(3\beta+2\alpha)-\cos(\beta-2\alpha)]\\
&\Leftrightarrow&\cos(3\alpha+2\beta)-\cos(3\beta+2\alpha)=\cos(\alpha-2\beta)-\cos(\beta-2\alpha)\\
&\Leftrightarrow&-2\sin\frac{5(\alpha+\beta)}{2}\sin\frac{\alpha-\beta}{2}=-2\sin\frac{3(\alpha+\beta)}{2}\sin\frac{-\alpha-\beta}{2}\\
&\Leftrightarrow&\sin\frac{5(\alpha+\beta)}{2}\sin\frac{\alpha-\beta}{2}=-\sin\frac{3(\alpha-\beta)}{2}\sin\frac{\alpha+\beta}{2}\\
&\Leftrightarrow&\sin\frac{\alpha-\beta}{2}\left[16\sin^5 \frac{\alpha+\beta}{2}-20\sin^3 \frac{\alpha+\beta}{2}+5\sin \frac{\alpha+\beta}{2}\right]\\
&&=-\sin\frac{\alpha-\beta}{2}\left[3-4\sin^2\frac{\alpha-\beta}{2}\right]\sin\frac{\alpha+\beta}{2}.
\end{eqnarray*}
Thus we have 
Case 1: $\sin\frac{\alpha-\beta}{2}=0$. Since $\alpha,\beta$ are acute angles, so $\alpha=\beta$.
Case 2: $16\sin^5 \frac{\alpha+\beta}{2}-20\sin^3 \frac{\alpha+\beta}{2}+5\sin \frac{\alpha+\beta}{2}=-\left[3-4\sin^2\frac{\alpha-\beta}{2}\right]\sin\frac{\alpha+\beta}{2}$. Since $\sin\frac{\alpha+\beta}{2}\neq 0$, we obtain
$$ 4\sin^4 \frac{\alpha+\beta}{2}-5\sin^2 \frac{\alpha+\beta}{2}+2=\sin^2\frac{\alpha-\beta}{2}.$$
Letting $x=\sin^2\frac{\alpha+\beta}{2}$, we have
$$ 4x^2-5x+1=-1+\sin^2\frac{\alpha-\beta}{2}$$
or
$$ (1-x)(1-4x)=-1+\sin^2\frac{\alpha-\beta}{2}. $$
Note $x\in[0,1]$. If $\frac{1}{4}\le x\le 1$, then $1-4x\le 0$ and hence
$$ (1-x)(1-4x)=-(1-x)(4x-1)=-\frac{1}{4}(4-4x)(4x-1)\ge -\frac{1}{4}\frac{9}{4}=-\frac{9}{16} $$
from which, we have $-1+\sin^2\frac{\alpha-\beta}{2}\ge -\frac{9}{16}$ or $\sin^2\frac{\alpha-\beta}{2}\ge \frac{7}{16}>0$. This implies $\alpha\neq \beta$. 
If $0\le x<\frac{1}{4}$, we have $(1-x)(1-4x)> 0$ and $-1+\sin^2\frac{\alpha-\beta}{2}\le 0$ and hence this will never occur.
Thus only Case 1 holds or $\alpha=\beta$.
