已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?已知(π/3+α)=-1/3,则cos(7π/6-α)=?

已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?已知(π/3+α)=-1/3,则cos(7π/6-α)=?
已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?
已知(π/3+α)=-1/3,则cos(7π/6-α)=?

已知cos(π/6+α)=1/2,则sin(2π/3+α)=?已知cos(π/2+α)=2/3,且α∈(-π/2,0),那么tan(3π/2+α)=?已知(π/3+α)=-1/3,则cos(7π/6-α)=?
1.
cos(π/6+α)=-sin(π/6+α+π/2)=-sin(2π/3+α)=1/2,
所以
sin(2π/3+α)=-1/2,
2.因为α∈(-π/2,0)
所以
π/2+α∈(0,π/2)
cos(π/2+α)=2/3
sin(α)=-2/3
cos(α)=√1-sin²α=√5/3
tan(3π/2+α)
=-cot(α)
=-cosα/sinα=-(√5/3)/(-2/3)=√5/2

sin(2π/3+α)=sin(π/2+π/6+α)=cos(π/6+α)=1/2