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)

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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)

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