(sin(180°+2x))/(1+cos2x)*(cos2x0/(cos(90°+x))=
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(sin(180°+2x))/(1+cos2x)*(cos2x0/(cos(90°+x))=
(sin(180°+2x))/(1+cos2x)*(cos2x0/(cos(90°+2x))
=[-sin2x/(1+cos2x)][cos2x/(-sinx)]
=(sin2xcos2x)/(sinx(1+cos2x))
=(sin2xcos2x)/[2snxcos²x]
=(sin2xcos2x)/sin2xcosx
=cos2x/cosx
再问: 不对吧,答案里没有
再答: (sin(180°+2x))/(1+cos2x)*(cos2x0/(cos(90°+2x))
=[-sin2x/(1+cos2x)][cos2x/(-sinx)]
=(sin2xcos2x)/(sinx(1+cos2x))
=(sin2xcos2x)/[2snxcos²x]
=(sin2xcos2x)/sin2xcosx
=cos2x/cosx
=(2cos²x-1)/cosx
=2cosx-secx
再问: 这个也没有,A.-sinx B.-cosx C.sinx D.cosx
((sin(180°+2x))/(1+cos2x))*((cosx平方/(cos(90°+x)))=
再答: (sin(180°+2x))/(1+cos2x)*(cos²x)/(cos(90°+2x))
=[-sin2x/(1+cos2x)][cos²x/(-sinx)]
=(sin2xcos²x)/(sinx(1+cos2x))
=(sin2xcos²x)/[2snxcos²x]
=(sin2xcos²x)/sin2xcosx
=cos²x/cosx
=cosx
=[-sin2x/(1+cos2x)][cos2x/(-sinx)]
=(sin2xcos2x)/(sinx(1+cos2x))
=(sin2xcos2x)/[2snxcos²x]
=(sin2xcos2x)/sin2xcosx
=cos2x/cosx
再问: 不对吧,答案里没有
再答: (sin(180°+2x))/(1+cos2x)*(cos2x0/(cos(90°+2x))
=[-sin2x/(1+cos2x)][cos2x/(-sinx)]
=(sin2xcos2x)/(sinx(1+cos2x))
=(sin2xcos2x)/[2snxcos²x]
=(sin2xcos2x)/sin2xcosx
=cos2x/cosx
=(2cos²x-1)/cosx
=2cosx-secx
再问: 这个也没有,A.-sinx B.-cosx C.sinx D.cosx
((sin(180°+2x))/(1+cos2x))*((cosx平方/(cos(90°+x)))=
再答: (sin(180°+2x))/(1+cos2x)*(cos²x)/(cos(90°+2x))
=[-sin2x/(1+cos2x)][cos²x/(-sinx)]
=(sin2xcos²x)/(sinx(1+cos2x))
=(sin2xcos²x)/[2snxcos²x]
=(sin2xcos²x)/sin2xcosx
=cos²x/cosx
=cosx
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