已知tan(π+α)=1/2,α∈（-π,0）,求tan(π-α),sin(α-π)的值
tan(π-α)=tan(-π-α)=-tan(π+α)=-1/2
2sin(π-α)=-cos(π-α)
[sin(π-α)]^2+[cos(π-α)]^2=1
sin(π-α)