已知cosA=(acosB-b)/(a-bcosB),求证(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
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已知cosA=(acosB-b)/(a-bcosB),求证(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
已知cosA=(acosB-b)/(a-bcosB),求证(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
已知cosA=(acosB-b)/(a-bcosB),求证(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
由万能公式
cosA=(1-(tanA/2)^2)/(1+(tanA/2)^2)
得
(tanA/2)^2=(1-cosA)/(1+cosA)
同理
(tanB/2)^2=(1-cosB)/(1+cosB)
所以
(tanA/2)^2/(tanB/2)^2=(1-cosAcosB-cosA+cosB)/(1-cosAcosB+cosA-cosB)……(1)
而已知
cosA=(acosB-b)/(a-bcosB)
即
a(cosB-cosA)=b(1-cosAcosB)
代入(1)式
可得
(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
把cos全化成1减tan平方再除1减tan
由万能公式
cosA=(1-(tanA/2)^2)/(1+(tanA/2)^2)
得
(tanA/2)^2=(1-cosA)/(1+cosA)
同理
(tanB/2)^2=(1-cosB)/(1+cosB)
所以
(tanA/2)^2/(tanB/2)^2=(1-cosAcosB-cosA+cosB)/(1-cosAcosB+cosA-cos...
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由万能公式
cosA=(1-(tanA/2)^2)/(1+(tanA/2)^2)
得
(tanA/2)^2=(1-cosA)/(1+cosA)
同理
(tanB/2)^2=(1-cosB)/(1+cosB)
所以
(tanA/2)^2/(tanB/2)^2=(1-cosAcosB-cosA+cosB)/(1-cosAcosB+cosA-cosB)……(1)
而已知
cosA=(acosB-b)/(a-bcosB)
即
a(cosB-cosA)=b(1-cosAcosB)
代入(1)式
可得
(tanA/2)^2/(tanB/2)^2=(a+b)/(a-b)
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