Double angle Identities

用戶221095

2008-01-01 2:21 pm

1) cosx-sinx/cosx+sinx=sec2x-tan2x

2)1-cos2x+sin2x/1+cos2x+sin2x=tanx

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回答 (1)

myisland8132

2008-01-01 3:58 pm

✔ 最佳答案

1) cosx-sinx/cosx+sinx= sec2x-tan2x
LHS
=(cosx-sinx)/(cosx+sinx)
=(cosx-sinx)(cosx-sinx)/(cosx+sinx)(cosx-sinx)
=(cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)
=(1-sin2x)/cos2x
=sec2x-tan2x
=RHS
2)1-cos2x+sin2x/1+co s2x+sin2x=tanx
LHS
=(1-cos2x+sin2x)/(1+co s2x+sin2x)
=(cos^2x+sin^2x-cos2x+sin2x)/(cos^2x+sin^2x+co s2x+sin2x)
=[(cosx+sinx)^2-cos2x]/[(cosx+sinx)^2+cos2x]
=[(cosx+sinx)^2-(cosx+sinx)(cosx-sinx)]/[(cosx+sinx)^2+(cosx+sinx)(cosx-sinx)]
=[(cosx+sinx)(cosx+sinx-cosx+sinx)]/[(cosx+sinx)(cosx+sinx+cosx-sinx)]
=[(cosx+sinx)(2sinx)]/[(cosx+sinx)(2cosx)]
=tanx
=RHS

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