Prove $x=y$, given $\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$ Given $x,y\in(0,\frac{\pi}{2}]$, $\alpha>0$ is a constant satisfying $0<x+y+\alpha<\pi$,
Also it is known that
$$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
How can we prove $x=y$ "elegantly"? I am stuck on this problem for several days. Please help.
 A: Put $x=x+\frac{\alpha}{2}-\frac{\alpha}{2}$ to get $$\sin(x)=\sin(x+\frac{\alpha}{2})\cos(\frac{\alpha}{2})-\sin(\frac{\alpha}{2})\cos(x+\frac{\alpha}{2})$$ and put $x+\alpha=x+\frac{\alpha}{2}+\frac{\alpha}{2}$ to get
$$\sin(x+\alpha)=\sin(x+\frac{\alpha}{2})\cos(\frac{\alpha}{2})+\sin(\frac{\alpha}{2})\cos(x+\frac{\alpha}{2})$$
Now with $u=\sin(x+\frac{\alpha}{2})$ we have:
$$\frac{1}{\sin(x)}+\frac{1}{\sin(x+\alpha)}=2\cos(\frac{\alpha}{2})\frac{u}{u^2-(\sin(\frac{\alpha}{2}))^2}$$
Do the same with the other expression with $v=\sin(y+\frac{\alpha}{2})$. You get easily $u=v$ or $uv=-(\sin(\frac{\alpha}{2}))^2$, and it is easy, using $x+y+\alpha<\pi$, to finish. 
A: $$\frac{1}{\sin x}+\frac{1}{\sin(x+\alpha)}=\frac{1}{\sin y}+\frac{1}{\sin(y+\alpha)}$$
or, $$\frac{1}{\sin x}-\frac{1}{\sin y}=\frac{1}{\sin(y+\alpha)}-\frac{1}{\sin(x+\alpha)}$$
or, $$\frac{\sin y-\sin x}{\sin x \sin y}=\frac{\sin(x+\alpha)-\sin(y+\alpha)}{\sin(y+\alpha)\sin(x+\alpha)}$$
or, $$-\frac{2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})}{\sin x \sin y}=\frac{2\cos(\frac{x+y}{2}+\alpha)\sin(\frac{x-y}{2})}{\sin(y+\alpha)\sin(x+\alpha)}$$
or, $$2\sin(\frac{x-y}{2})\left(\frac{\cos(\frac{x+y}{2})}{\sin x \sin y}+\frac{\cos(\frac{x+y}{2}+\alpha)}{\sin(y+\alpha)\sin(x+\alpha)}\right)=0$$
It is very easy to say $x=y$ from here. I hope you can do this last step.
