# If xcostheta=ycos(theta+(2pi)/3)=zcos(theta+(4pi)/3)=1/k calculate xy+yz+zx= ?

Dec 21, 2016

Calling $\lambda = x \cos \left(\theta\right)$ we have

${\left(x + y + z\right)}^{2} = {\left(\frac{\lambda}{\cos} \left(\theta\right) + \frac{\lambda}{\cos} \left(\theta + \frac{2 \pi}{3}\right) + \frac{\lambda}{\cos} \left(\theta + \frac{4 \pi}{3}\right)\right)}^{2} = \frac{9 {\lambda}^{2}}{\cos} {\left(3 \theta\right)}^{2}$

Now making ${x}^{2} + {y}^{2} + {z}^{2}$ we arrive at the same result. Considering

${\left(x + y + z\right)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2 \left(x y + x z + y z\right)$

we conclude

$x y + x z + y z = 0$

Dec 21, 2016

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

So

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

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

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

Now

Given
x,y,z are non-zero real numbers

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

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

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

$= k \left(\cos \theta - 2 \cdot \cos \theta \times \frac{1}{2}\right)$

$= k \left(\cos \theta - \cos \theta\right) = k \times 0 = 0$

$\implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

$\implies \frac{x y + y z + z x}{x y z} = 0$

$\implies x y + y z + z x = 0$