sin(3θ)=sin(2θ)cos(θ)+cos(2θ)sin(θ)=
2sin(θ)cos2(θ)+(1-2sin2(θ))sin(θ)=2sin(θ)(1-sin2(θ))+(1-2sin2(θ))sin(θ)=
2sin(θ)-2sin3(θ)+sin(θ)-2sin3(θ)=3sinθ-4sin3(θ).
sin(θ)=½(x+1/x), so:
sin(3θ)=3[½(x+1/x)]-4[½(x+1/x)]3,
sin(3θ)=3x/2+3/(2x)-½(x3+3x+3/x+1/x3),
sin(3θ)=3x/2+3/(2x)-x3/2-3x/2-3/(2x)-1/(2x3),
sin(3θ)=-½(x3+1/x3), so sin(3θ)+½(x3+1/x3)=0 QED