如题,求证:sin2x/ [(sinx+cosx-1)(sinx+1-cosx)] =cot(x/2)
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如题,
求证:sin2x/ [(sinx+cosx-1)(sinx+1-cosx)] =cot(x/2)
求证:sin2x/ [(sinx+cosx-1)(sinx+1-cosx)] =cot(x/2)
sin2x/ [(sinx+cosx-1)(sinx+1-cosx)]
=sin2x/ [(sinx-(1-cosx)(sinx+1-cosx)]
=sin2x/ [(sin^2x-(1-cosx)^2]
=sin2x/ [sin^2x-1-cos^2x+2cosx]
=sin2x/ [sin^2x-(sin^2x+cos^2x)-cos^2x+2cosx]
=sin2x/ [-2cos^2x+2cosx]
=2sinxcosx/2cosx(1-cosx)
=sinx/(1-cosx)
=cot(x/2)
=sin2x/ [(sinx-(1-cosx)(sinx+1-cosx)]
=sin2x/ [(sin^2x-(1-cosx)^2]
=sin2x/ [sin^2x-1-cos^2x+2cosx]
=sin2x/ [sin^2x-(sin^2x+cos^2x)-cos^2x+2cosx]
=sin2x/ [-2cos^2x+2cosx]
=2sinxcosx/2cosx(1-cosx)
=sinx/(1-cosx)
=cot(x/2)
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