求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2).
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求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2).
[sin 2x /(1-cos 2x)]·[sin x /(1+sin x)]
=2sinx*cosx*sinx/[2(sinx)^2*(1+sinx)]
=cosx/(1+sinx)
=[(cosx/2)^2-(sinx/2)^2]/[(cosx/2)^2+(sinx/2)^2+2cosx/2*sinx/2]
=[(cosx/2+sinx/2)*(cosx/2-sinx/2)]/[(cosx/2+sinx/2)^2
=(cosx/2-sinx/2)]/(cosx/2+sinx/2)
=(1-tanx/2)/(1+tanx/2)
=(tanπ/4-tanx/2)/(1+tanπ/4*tanx/2)
=tan (π/4-x/2)
=2sinx*cosx*sinx/[2(sinx)^2*(1+sinx)]
=cosx/(1+sinx)
=[(cosx/2)^2-(sinx/2)^2]/[(cosx/2)^2+(sinx/2)^2+2cosx/2*sinx/2]
=[(cosx/2+sinx/2)*(cosx/2-sinx/2)]/[(cosx/2+sinx/2)^2
=(cosx/2-sinx/2)]/(cosx/2+sinx/2)
=(1-tanx/2)/(1+tanx/2)
=(tanπ/4-tanx/2)/(1+tanπ/4*tanx/2)
=tan (π/4-x/2)
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