Question

If $x,y,z$ are non zero real numbers and if $x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})$ for some $\theta \in R$, then show that $xy+yz+zx=0$

Answer 1

$x\cos \theta =y\cos (\theta +\frac{2\pi }{3})=z\cos (\theta +\frac{4\pi }{3})=\frac{1}{k}$

$\frac{1}{x}=k\cos \theta$

$\frac{1}{y}=k\cos (\theta +\frac{2\pi }{3})$

$\frac{1}{z}=k\cos (\theta +\frac{4\pi }{3})$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k\{\cos \theta +\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{4\pi }{3}\left)\right\}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k\{\cos \theta +\cos (\theta +\frac{4\pi }{3})+\cos (\theta +\frac{2\pi }{3}\left)\right\}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k\left\{2\cos \right(\theta +\frac{2\pi }{3}\left)\cos \right(\frac{2\pi }{3})+\cos (\theta +\frac{2\pi }{3}\left)\right\}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k\left\{2\right(-\frac{1}{2}\left)\cos \right(\theta +\frac{2\pi }{3})+\cos (\theta +\frac{2\pi }{3}\left)\right\}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k\{-\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{2\pi }{3}\left)\right\}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0⟹\frac{yz+xz+xy}{xyz}=0$

$xy+yz+xz=0$ (Proved)