证明(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2

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证明(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2
证明(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2

证明(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2
(sinx+cosx-1)(sinx-cosx+1)/sin2x
=[sinx+(cosx-1)][sinx-(cosx-1)]/sin2x
=[(sinx)^2-(cosx-1)^2]/sin2x
=[(sinx)^2-(cosx)^2-1+2cosx]/sin2x
={(sinx)^2-(cosx)^2-[(sinx)^2+(cosx)^2]+2cosx}/sin2x
=[2cosx-2(cosx)^2]/sin2x
=2cosx(1-cosx)/2sinxcosx
=(1-cosx)/sinx
=tanx/2
得证

(sinx+cosx-1)(sinx-cosx+1)/sin2x
=[sin^2x-(cosx-1)^2]/2sinxcosx
=(sin^2x-cos^2x+2cosx-1)/2sinxcosx
=(-2cos^2x+2cosx)/2sinxcosx
=(1-cosx)/sinx
tanx/2
=(sinx/2)/(cosx/2)
=[si...

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(sinx+cosx-1)(sinx-cosx+1)/sin2x
=[sin^2x-(cosx-1)^2]/2sinxcosx
=(sin^2x-cos^2x+2cosx-1)/2sinxcosx
=(-2cos^2x+2cosx)/2sinxcosx
=(1-cosx)/sinx
tanx/2
=(sinx/2)/(cosx/2)
=[sinx/(2cosx/2)]/(cosx/2)
=sinx/(2cos^2x/2)
=sinx/(1+cosx)
=[sinx(1-cosx)]/[(1+cosx)(1-cosx)]
=sinx(1-cosx)/sin^2x
=(1-cosx)/sinx
左=右

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