Prove that $\frac{\sin^22\alpha}{\sin(2\alpha+\beta)}=\frac{\sin^22\beta}{\sin(2\beta+\alpha)}$ is true only for $\alpha=\beta$

Question

Interesting trigonometric equation showed up while I was trying to solve a geometry problem:

$$\frac{\sin^22\alpha}{\sin(2\alpha+\beta)}=\frac{\sin^22\beta}{\sin(2\beta+\alpha)}\tag{1}$$

...under condition that $\alpha,\beta$ are angles of a triangle. The trick is to show that (1) is true only in trivial case $\alpha=\beta$.

I have tried to prove it in a brute-force style by getting rid of fractions and by expanding everything that I could expand. But the computation proved to be messy and I was not patiet enough to bring it to any conclusion. Any ideas how to tackle this kind of problem?

Answer 1

If you rewrite your equation as $$\sin ^2(2 a) \sin (a+2 b)-\sin ^2(2 b) \sin (2 a+b)=0$$ expand the sines and simplify to get $$\sin (2 a-3 b)+\sin (3 a-2 b)-2 \sin (2 a+b)+2 \sin (a+2 b)-\sin (5 a+2 b)+\sin (2 a+5 b)=0$$