sin2x.sinx \1-cosx

y=(sin2x·sinx)/(1-cosx)=2sinxcosxsinx/(1-cosx)=2sin^2xcosx/(1-cosx)=2(1-cos^2x)sinx/(1-cosx)
y=(sin2x·sinx)/(1-cosx)=2(1+cosx)cosx=2cosx+2cos^2x=2(cosx+1/2)^2-1/2
所以值域为-1/2=