有2條maths我計唔到...help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2008-03-27 10:53 pm

條數要証明佢地係identitie:
no1
tanθsinθ tanθ-sinθ
------------- = --------------
tanθ+sinθ tanθsinθ
no2
cos^2θ-sin^2θ 1-tanθ
------------------ = ----------------
1+2sinθcosθ 1+tanθ

我証明唔到....有冇高手?

回答 (1)

HaHa

2008-03-27 11:35 pm

✔ 最佳答案

1)
tanθsinθ/(tanθ+sinθ)
=[tanθsinθ/(tanθ+sinθ)][(tanθ- sinθ)/(tanθ-sinθ)
=tanθsinθ(tanθ- sinθ)/(tan^2θ-sin^θ)
=tanθsinθ(tanθ-sinθ)/[tan^2θ(1-cos^2θ)
=tanθsinθ(tanθ-sinθ)/tan^2θsin2θ
=(tanθ-sinθ)/(tanθsinθ)

2)(1-tanθ)/(1+tanθ)
=[(1-tanθ)/(1+tanθ)](1+tanθ)/(1+tanθ)
=(1-tan^2θ)/(1+tanθ)^2
=(1/cos^2θ)(cos^2θ-sin^2θ)／(1/cos^2θ)(cosθ+sinθ)^2
=(cos^2θ-sin^2θ)／(cos^2θ+2cosθsinθ+sin^2θ)
=(cos^2θ-sin^2θ)／(1+2cosθsinθ)

希望幫到你！

2008-03-27 15:37:11 補充：
對唔住！打漏了＾第一條的sin2θ應是sin^2θ

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