2
$\begingroup$

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

$\endgroup$
1
  • 3
    $\begingroup$ You have 12 variables and 8 equations. The degree of freedom of an octahedron is $\dim SO(3)$ which is 3. So I suspect one equation is missing? $\endgroup$ – John Jiang Jan 3 '16 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.