Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin. 

It's just trigonometric manipulation.
$$\frac{\sin(a-b)}{\sin(a+b)}+\frac{\sin(b-c)}{\sin(b+c)}=\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\sin(b+c)}$$
We can now use $\sin x\sin y=\frac{1}{2}(\cos(x-y)-\cos(x+y))$ to gwtite the above expression as 
$$\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\sin(b+c)}=\frac{\cos(a-2b-c)-\cos(a+c)+\cos(a+c)-\cos(a+2b-c)}{2\sin(a+b)\sin(b+c)}$$
The use $\cos x-\cos y=2\sin\frac{x-y}{2}\sin\frac{x+y}{2}$ to get
$$\frac{\sin(2b)\sin(a-c)}{\sin(a+b)\sin(b+c)}$$
If we go back to the given equation, we see that we have $\sin(a-c)$ as a common term, so $a-c=0$ is one of the solutions. So the only thing left to do is to show that $$\sin(2b)\sin(c+a)+\sin(a+b)\sin(b+c)=0$$ contains the other two solutions.
Using the same identity as above we get $$\sin(2b)\sin(c+a)-\sin(a+b)\sin(b+c)=\frac{1}{2}(\cos(a-2b+c)-\cos(a+2b+c)-\cos(a-c)+\cos(a+2b+c))=-\sin(a-b)\sin(b-c)=0$$ 
