Could somebody help me to prove the following?
$$\sum_{k=1}^6 \cos(2 \theta_k) (\cos(2\phi_k)-1))=0$$ $$\sum_{k=1}^6 \sin(2 \theta_k) (\cos(2\phi_k)-1))=0$$ $$\sum_{k=1}^6 \cos (\phi_k)=0$$ $$\sum_{k=1}^6 \cos (2 \phi_k)=-2$$ $$\sum_{k=1}^6 \cos (\theta_k) \sin(\phi_k)=0$$ $$\sum_{k=1}^6 \sin (\theta_k) \sin(\phi_k)=0$$ $$\sum_{k=1}^6 \cos(\theta_k) \sin(2 \phi_k)=0$$ $$\sum_{k=1}^6 \sin(\theta_k) \sin(2 \phi_k)=0$$ if and only if $\{(\cos (\theta_k)\sin(\phi_k), \sin (\theta_k) \sin(\phi_k), \cos(\phi_k) \}_{k\in \{1,2,\ldots,6\}}$ forms the vertices of a regular octahedron.
Actually,the statement that if $\{(\cos (\theta_k)\sin(\phi_k), \sin (\theta_k) \sin(\phi_k), \cos(\phi_k) \}_{k\in \{1,2,\ldots,6\}}$ forms the vertices of a regular octahedron, then the equations (1)-(8) hold is quite clear. The part that I have the problem with is the inverse part.
Thank you so much.
Masih