tanx=-2,π/2

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tanx=-2,π/2tanx=-2,π/2tanx=-2,π/2sin(x+π/6)(sin2x+sin^2x)/(1-cos2x)=sin(x+π/6)(2sinxcosx+sin^2x)/2si

tanx=-2,π/2
tanx=-2,π/2

tanx=-2,π/2
sin(x+π/6) (sin2x+sin^2x)/(1-cos2x)
=sin(x+π/6)(2sinxcosx+sin^2x)/2sin^2x
=[1/2 +cosx/sinx]sin(x+π/6)
=[1/2+1/tanx]sin(x+π/6)
=0
很高兴为你解决问题

0

tanx=-2,π/2sinx=2/根号5,cosx=-1/根号5;
sin2x=-4/5;
sin2x+sin^2x=0
sin(x+π/6)(sin2x+sin^2x)/(1-cos2x)=0