直角三角形において、sin[stuv]^2+cos[stuv]^2=1

(t,u)(s,v)+(s,t)(u,v)=(s,u)(t,v)

(t,u)(s,v)+(u,s)(t,v)(s,t)(u,v)=0

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辺AiAj=[-1,i,j,4]

AiAjに対向する辺の二面角=[i-1,i,j,j+1]

それ以外

(-1,0)(0,1)(1,2)(2,3)(3,4)

(-1,1)(0,2)(1,3)(2,4)

(-1,2)(0,3)(1,4)

(-1,3)(0,4)

(-1,4)

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1行目を1とおくと

(secα)^2=(-1,1)(0,2)

(secβ)^2=(0,2)(1,3)

(secγ)^2=(1,3)(2,4)

(tanα)^2= (-1,2)

(tanβ)^2= (0,3)

(tanγ)^2= (1,4)

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