化简:(1-cosx-sinx)/(1+cosx-sinx)(2cos10°-sin20°)/cos20°求值域:y=cos2x+2sinx能做几道是几道,o(∩_∩)o...
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化简:(1-cosx-sinx)/(1+cosx-sinx)(2cos10°-sin20°)/cos20°求值域:y=cos2x+2sinx能做几道是几道,o(∩_∩)o...
化简:(1-cosx-sinx)/(1+cosx-sinx)
(2cos10°-sin20°)/cos20°
求值域:y=cos2x+2sinx
能做几道是几道,o(∩_∩)o...
化简:(1-cosx-sinx)/(1+cosx-sinx)(2cos10°-sin20°)/cos20°求值域:y=cos2x+2sinx能做几道是几道,o(∩_∩)o...
第1题
分子=(1-cosx-sinx)=[sin^2(x/2)+cos^2(x/2)]-[cos^2(x/2)-sin^2(x/2)]-2sin(x/2)cos(x/2)=2sin^2(x/2)-2sin(x/2)cos(x/2)=2sin(x/2)[sin(x/2)-cos(x/2)]
分母=(1+cosx-sinx) =[sin^2(x/2)+cos^2(x/2)]+[cos^2(x/2)-sin^2(x/2)]-2sin(x/2)cos(x/2)=2cos^2(x/2)-2sin(x/2)cos(x/2)=-2cos(x/2)[sin(x/2)-cos(x/2)]
(1-cosx-sinx)/(1+cosx-sinx)
=2sin(x/2)[sin(x/2)-cos(x/2)]/-2cos(x/2)[sin(x/2)-cos(x/2)]
=-tan(x/2)
第2题
第3题
y=cos2x+2sinx
y=-2(sinx+1/2)^2+3/2
当sinx=-1时,y=1
当sinx=1是,y=-3
所以值域为[-3,1]