Calculus sine proof

Question

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.

Answer 1

$$\frac{\sin(a-b)}{\sin(a+b)}=\frac{\alpha/\beta-\beta/\alpha} {\alpha\beta-1/(\alpha\beta)}=\frac{\alpha^2-\beta^2}{\alpha^2\beta^2-1}$$ where $\alpha=\exp(ia)$ and $\beta=\exp(ib)$. Set $\gamma=\exp(ic)$. Then $$(\alpha^2-\beta^2)(\alpha^2\gamma^2-1)(\beta^2\gamma^2-1) +(\beta^2-\gamma^2)(\alpha^2\beta^2-1)(\alpha^2\gamma^2-1) +(\gamma^2-\alpha^2)(\alpha^2\beta^2-1)(\beta^2\gamma^2-1) =0.$$ This simplifies to $$(\alpha^4-\beta^4)\gamma^2+(\beta^4-\gamma^4)\alpha^2 +(\gamma^4-\alpha^4)\beta^2=0.$$ But that factors as $$(\alpha^2-\beta^2)(\beta^2-\gamma^2)(\gamma^2-\alpha^2)=0.$$ If say $\alpha^2=\beta^2$ then $a=b$, as $a$ and $b$ are acute angles.

Answer 2

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 $$\sin(2b)\sin(c+a)-\sin(a+b)\sin(b+c)=\frac{1}{2}(\cos(a-2b+c)-\cos(a+2b+c)-\cos(a-c)+\cos(a+2b+c))=-\sin(a-b)\sin(b-c)=0$$

Answer 3

Let $\tan{a}=x$, $\tan{b}=y$ and $\tan{c}=z$.

Thus, $$0=\sum_{cyc}\frac{\sin(a-b)}{\sin(a+b)}=\sum_{cyc}\frac{x-y}{x+y}=\frac{\sum\limits_{cyc}(x-y)(x+z)(y+z)}{\prod\limits_{cyc}(x+y)}=$$ $$=\frac{\sum\limits_{cyc}(x-y)(z^2+xy+xz+yz)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x-y)z^2}{\prod\limits_{cyc}(x+y)}=$$ $$=\frac{\sum\limits_{cyc}(x^2y-x^2z)}{\prod\limits_{cyc}(x+y)}=\frac{(x-y)(x-z)(y-z)}{(x+y)(x+z)(y+z)},$$ which gives $$(a-b)(a-c)(b-c)=0$$ and we are done!