sin(3θ)=sin(2θ+θ)=(2sin(θ)cosθ)cos(θ)+(1-2sin2(θ))sin(θ)=
2sin(θ)cos2(θ)+sin(θ)-2sin3(θ)=2sin(θ)(1-sin2(θ))+sin(θ)-2sin3(θ)=
3sin(θ)-4sin3(θ)=sin(θ)(3-4sin2(θ))=½(x+1/x)(3-4(¼(x2+2+1/x2)))=
½(x+1/x)(1-x2-1/x2)=½(x-x3-1/x+1/x-x-1/x3)=-½(x3+1/x3).
Therefore sin(3θ)=-½(x3+1/x3), and sin(3θ)+½(x3+1/x3)=0 QED