If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$
This problem was bothering me for a while, and I finally could not solve it. I tried taking the whole equation as $k$ but the calculation was a mess. Would someone please help me to find a solution using a simpler approach?
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Consider (first & second terms) $$x\sin A=y\sin\left(A+\frac{2\pi}{3}\right)$$ $$x\sin A=y\left(\sin A\cos \frac{2\pi}{3}+\cos A\sin\frac{2\pi}{3}\right)$$
$$x=y\left(\frac{\sin A\cos \frac{2\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{2\pi}{3}}{\sin A}\right)$$
$$x=y\left(-\frac12+\cot A \frac{\sqrt3}{2}\right)$$ $$\cot A=\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)\tag 1$$
Similarly, consider (first & third terms) $$x\sin A=z\sin\left(A+\frac{4\pi}{3}\right)$$ $$x\sin A=z\left(\sin A\cos \frac{4\pi}{3}+\cos A\sin\frac{4\pi}{3}\right)$$ $$x=z\left(\frac{\sin A\cos \frac{4\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{4\pi}{3}}{\sin A}\right)$$
$$x=z\left(-\frac12+\cot A \left(-\frac{\sqrt3}{2}\right)\right)$$ $$\cot A=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)\tag 2$$ Equating (1) & (2), we get $$\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)$$ $$\frac xy+\frac xz+1=0$$
Hope above is the required relation among $x, y$ & $z$
Use $$\sin{A}+\sin(A+120^{\circ})+\sin(A+240^{\circ})=0.$$ Indeed, let $$x\sin{A}=y\sin(A+120^{\circ})=z\sin(A+240^{\circ})=k.$$
This, for $xyz\neq0$ and $k\neq0$ we obtain: $$\frac{k}{x}+\frac{k}{y}+\frac{k}{z}=0$$ or $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.$$