若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=

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若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=

若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=
若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=

若tan(π+α)=m,3π/2<α<2π,则sin(π-α)=
3π/2<α<2π
5π/2<α+π<3π
sin(α+π)>0,tan(α+π)<0,m<0
sin(α+π)=-tan(α+π)/√[1+tan^2(α+π)]
=-m/√(1+m^2)
=-sinα
sinα=m/√(1+m^2)<0
-π<π-α<-π/2
sin(π-α)<0
=sina=m/√(1+m^2)