If cos x + cos y + cos z = 0 = sin x + sin y+ sinz, then possible value of cos{(x-y)/2} is?​  

Dear student

We have,
$$\begin{gathered} \mathrm{cosx}+\mathrm{cosy}+\mathrm{cosz}=0 \\ \Rightarrow \mathrm{cosx}+\mathrm{cosy}=-\mathrm{cosz} ....\left(1\right) \\ \mathrm{and} \mathrm{sinx}+\sin +\mathrm{sinz}=0 \\ \Rightarrow \mathrm{sinx}+\mathrm{siny}=-\mathrm{sinz} .....\left(2\right) \end{gathered}$$
Squaring on both sides of equation (1) and (2), we get
$$\begin{gathered} {\left(\mathrm{cosx}+\mathrm{cosy}\right)}^2={\cos }^2\mathrm{z} \\ \Rightarrow {\cos }^2\mathrm{x}+{\cos }^2\mathrm{y}+2\mathrm{cosx} \mathrm{cosy}={\cos }^2\mathrm{z} ... \left(3\right) \\ \mathrm{and} (\mathrm{sinx}+\mathrm{siny}{)}^2={\sin }^2\mathrm{z} \\ \Rightarrow {\sin }^2\mathrm{x}+{\sin }^2\mathrm{y}+2\mathrm{sinx} \mathrm{siny}={\sin }^2\mathrm{z} .... \left(4\right) \end{gathered}$$

adding eqs. (3) and (4),we get

$$\begin{gathered} {\cos }^2\mathrm{x}+{\cos }^2\mathrm{y}+2\mathrm{cosx} \mathrm{cosy}+{\sin }^2\mathrm{x}+{\sin }^2\mathrm{y}+2\mathrm{sinx} \mathrm{siny}={\cos }^2\mathrm{z}+{\sin }^2\mathrm{z} \\ \Rightarrow 1+2\mathrm{cosx} \mathrm{cosy}+1+2\mathrm{sinx} \mathrm{siny}=1 \\ \Rightarrow 2(\mathrm{cosx} \mathrm{cosy}+\mathrm{sinx} \mathrm{siny})=-1 \\ \Rightarrow 2\cos \left(\mathrm{x}-\mathrm{y}\right)=-1 \\ \Rightarrow \cos \left(\mathrm{x}-\mathrm{y}\right)=\frac{-1}{2} .....\left(5\right) \end{gathered}$$

Since we have to find the value of $\cos \left(\frac{x-y}{2}\right)$.
$\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(\mathrm{x}-\mathrm{y}\right)}{2}} = \pm \sqrt{\frac{1 -{\displaystyle \frac{1}{2}}}{2}} = \pm \left(\frac{1}{2}\right)$
Regards