Sines,cosines and zeros

In this question we need to use a few formulas

1)(SinH)^2+(cosH)^2=1

2)cos(H-K)=sinHcosK+cosHsinK

sinX+sinY+sinZ=0

SinX+sinY=-sinZ

(SinX+sinY)^2=(sinZ)^2

cosX+cosY+cosZ=0

cosX+cosY=-cosZ

(cosX+cosY)^2=(cosZ)^2

Add them up

(SinX+sinY)^2+(cosX+cosY)^2 =(cosZ)^2+(sinZ)^2

(SinX+sinY)^2+(cosX+cosY)^2=1

(SinX)^2+(sinY)^2+2sinXsinY+(cosX)^2+(cosY)^2+2cosXcosY=1

1+1+2sinXsinY+2cosXcosY=1

Cos(X-Y)=-0.5

The general solution for $\begin{cases} \sin x + \sin y + \sin z = 0 \\ \cos x + \cos y + \cos z = 0 \end{cases}$ is $(x, y, z) = \left(\theta - \frac {2\pi}3, \theta, \theta + \frac {2\pi}3\right)$ for all real $\theta$ , where $x$ , $y$ , and $z$ can take anyone of the three values. The proof is as follows:

$\begin{cases} \sin x + \sin y + \sin z & = {\color{#3D99F6}\sin \left(\theta - \frac {2\pi}3\right)} + \sin \theta + {\color{#3D99F6}\sin \left(\theta + \frac {2\pi}3\right)} = {\color{#3D99F6}2 \sin \theta \cos \frac {2\pi}3} + \sin \theta = - \sin \theta + \sin \theta = 0 \\ \cos x + \cos y + \cos z & = {\color{#3D99F6}\cos \left(\theta - \frac {2\pi}3\right)} + \cos \theta + {\color{#3D99F6}\cos \left(\theta + \frac {2\pi}3\right)} = {\color{#3D99F6}2 \cos \theta \cos \frac {2\pi}3} + \sin \theta = - \cos \theta + \cos \theta = 0 \end{cases}$

Therefore, $\cos (x-y) = \begin{cases} \cos \left(\theta - \frac {2\pi}3-\theta \right) = \cos \left(-\frac {2\pi}3\right) = - \frac 12 \\ \cos \left(\theta - \theta - \frac {2\pi}3 \right) = \cos \left(-\frac {2\pi}3\right) = - \frac 12 \\ \cos \left(\theta + \frac {2\pi}3-\theta + \frac {2\pi}3 \right) = \cos \left(\frac {4\pi}3\right) = - \frac 12 \end{cases} = \boxed{-0.5}$ .