若cos(π+α)=-1/2,3π/2

若cos(π+α)=-1/2,3π/2
若cos(π+α)=-1/2,3π/2

若cos(π+α)=-1/2,3π/2
cos(π+α)= sin(-π/2 -α)= -sin(π/2 +α)= -cos(-α)= -cosα= -1/2
则cosα＝1/2,
sin(2π-α)＝sin(-α)= -sinα
3π/2

根号3/2

3π/2<α<2π
sinα<0
cos(π+α)=-1/2
-cosα=-1/2
cosα=1/2
sinα=-√3/2
sin(2π-α)
=-sinα
=-(-√3/2)
=√3/2

∵cos(π+α)=-1/2
∴-cosα=-1/2
cosα=1/2
sin(2π-α)
=-sinα
=-(-)(1-(1/2)^2)^1/2
=2分支根号3
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