任意四面体的立体角

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对于任意一个四面体OABC,其中O,A,B,C分别为四面体的四个顶点,求证:从O点观察三角形ABC的立体角Ω的公式为$$\tan\left( \frac{\Omega}{4}\right)=\sqrt{ \tan \left( \frac{s}{2}\right) \tan \left( \frac{s - \alpha}{2}\right) \tan \left( \frac{s - \beta}{2}\right) \tan \left( \frac{s - \gamma}{2}\right)}$$其中α=∠BOC，β=∠AOC，γ=∠AOB均为平面角，$s = \frac {1}{2} (\alpha + \beta + \gamma).$
例如:对于正四面体,$α=β=γ=\fracπ3,s=\fracπ2,\Omega=4 \tan ^{-1}\left(\sqrt{\tan ^3\left(\frac{\pi }{12}\right)}\right)=3 \cos ^{-1}\left(\frac{1}{3}\right)-\pi=\cos^{-1}\left(\frac{23}{27}\right)≈0.55129$ rad.

hbghlyj

Todhunter, I. (1886). Spherical Trigonometry (5th ed.). Art.101—103
$$\tan {\tfrac {1}{4}}E={\sqrt {\tan {\tfrac {1}{2}}s\,\tan {\tfrac {1}{2}}(s-a)\,\tan {\tfrac {1}{2}}(s-b)\,\tan {\tfrac {1}{2}}(s-c)}}$$where $s=(a+b+c)/2$.