Question

sin(α+π/3)+sinα=-4根号3/5,则(α+2π/3)等于

Answer 1

sin(α+π/3)+sinα=2sin{[(α+π/3)+α]/2]}*cos{[(α+π/3)-α]/2}.
=2sin(α+π/6)cos(π/6).
=2sin(α+π/6)*(√3/2).
=√3sin(α+π/6).
∵sin(α+π/3)+sinα=-4√3/5.
∴√3sin(α+π/6)=-4√3/5
sin(α+π/6)=-4/5.
∵π/6=π/2-π/3.
∴sin(α+π/6)=sin(α+π/2-π/3).
sin(α+π/2-π/3)=cos(α-π/3).
即,cos(α-π/3)=-4/5..
∵α+2π/3=α+π-π/3.
cos(α+2π/3)=cos(π+α-π/3).
=-cos(α-π/3).
=-(-4/5).
=4/5).
∴α+2π/3=arccos(4/5)