■シュレーフリの公式と直角三角錐(その63)
1行目はすべて1
3行目は(tanα)^2・・・一意に決まる
2行目は2項の積が(secα)^2・・・一意に決まらないが、
(tanα)^2-(secα)^2=1
を満足する。どれかを一意に決めればすべて一意となる。
4行目は(tanα)^2(tanβ)^2-(secα)^2(secβ)^2=1?
(sinα)^2(sinβ)^2-1=(cosα)^2(cosβ)^2?
1/4{cos(α+β)-cos(α-β)}^2-1/4{cos(α+β)+cos(α-β)}^2=1?
-cos(α+β)cos(α-β)=1
-1/2{cos(2α)+cos(2β)}=1
5行目は4行目の2項の積が1であれば、0になる。
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a=[-1,0,1,4],(tana)^2=(0,1)(-1,4)/(-1,0)(1,4)
b=[-1,1,2,4],(tanb)^2=(1,2)(-1,4)/(-1,1)(2,4)
c=[-1,2,3,4],(tanc)^2=(2,3)(-1,4)/(-1,2)(3,4)
α=[-1,0,1,2],(tanα)^2=(0,1)(-1,2)/(-1,0)(1,2)
β=[0,1,2,3],(tanβ)^2=(1,2)(0,3)/(0,1)(2,3)
γ=[1,2,3,4],(tanγ)^2=(2,3)(1,4)/(1,2)(3,4)
(u,u+1)=1とおくと
a=[-1,0,1,4],(tana)^2=(-1,4)/(1,4)
b=[-1,1,2,4],(tanb)^2=(-1,4)/(-1,1)(2,4)
c=[-1,2,3,4],(tanc)^2=(-1,4)/(-1,2)
α=[-1,0,1,2],(tanα)^2=(-1,2)
β=[0,1,2,3],(tanβ)^2=(0,3)
γ=[1,2,3,4],(tanγ)^2=(1,4)
(-1,4)=0とおくとtanも消えて
a=[-1,0,1,4],(a)^2=1/(1,4)
b=[-1,1,2,4],(b)^2=1/(-1,1)(2,4)
c=[-1,2,3,4],(c)^2=1/(-1,2)
α=[-1,0,1,2],(tanα)^2=(-1,2)
β=[0,1,2,3],(tanβ)^2=(0,3)
γ=[1,2,3,4],(tanγ)^2=(1,4)
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4次元では
a=[-1,0,1,5],(tana)^2=(0,1)(-1,5)/(-1,0)(1,5)
b=[-1,1,2,5],(tanb)^2=(1,2)(-1,5)/(-1,1)(2,5)
c=[-1,2,3,5],(tanc)^2=(2,3)(-1,5)/(-1,2)(3,5)
d=[-1,3,4,5],(tand)^2=(3,4)(-1,5)/(-1,3)(4,5)
α=[-1,0,1,2],(tanα)^2=(0,1)(-1,2)/(-1,0)(1,2)
β=[0,1,2,3],(tanβ)^2=(1,2)(0,3)/(0,1)(2,3)
γ=[1,2,3,4],(tanγ)^2=(2,3)(1,4)/(1,2)(3,4)
δ=[2,3,4,5],(tanγ)^2=(3,4)(2,5)/(2,3)(4,5)
(u,u+1)=1とおくと
a=[-1,0,1,5],(tana)^2=(-1,5)/(1,5)
b=[-1,1,2,5],(tanb)^2=(-1,5)/(-1,1)(2,5)
c=[-1,2,3,5],(tanc)^2=(-1,5)/(-1,2)(3,5)
d=[-1,3,4,5],(tand)^2=(-1,5)/(-1,3)
α=[-1,0,1,2],(tanα)^2=(-1,2)
β=[0,1,2,3],(tanβ)^2=(0,3)
γ=[1,2,3,4],(tanγ)^2=(1,4)
δ=[2,3,4,5],(tanγ)^2=(2,5)
(-1,5)=0とおくとtanも消えて
a=[-1,0,1,5],(a)^2=1/(1,5)
b=[-1,1,2,5],(b)^2=1/(-1,1)(2,5)
c=[-1,2,3,5],(c)^2=1/(-1,2)(3,5)
d=[-1,3,4,5],(d)^2=1/(-1,3)
α=[-1,0,1,2],(tanα)^2=(-1,2)
β=[0,1,2,3],(tanβ)^2=(0,3)
γ=[1,2,3,4],(tanγ)^2=(1,4)
δ=[2,3,4,5],(tanγ)^2=(2,5)
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