試證〖sin〗^2 θ+〖sin〗^2 (θ+2π/3)+〖

Sol
A=Sin^2 θ+Sin^2 (θ+2π/3)+Sin^2 (θ+4π/3)
2A=2Sin^2 θ+2Sin^2 (θ+2π/3)+2Sin^2 (θ+4π/3)
=1－Cos(2θ)+1－Cos(2θ+4π/3)+1－Cos(2θ+8π/3)
=3－Cos(2θ)－Cos(2θ+4πI/3)－Cos(2θ+8π/3)
=3－Cos(2θ)－Cos(2θ)Cos(4π/3)+Sin(2θ)Sin(4π/3)
－Cos(2θ)Cos(8π/3)+Sin(2θ)Sin(8π/3)
=3－Cos(2θ)－Cos(2θ)*(－1/2)+Sin(2θ)*(－√3/2)
－Cos(2θ)*(－1/2)+Sin(2θ)*(√3/2)
=3－Cos(2θ)+Cos(2θ)*(1/2)+Cos(2θ)*(1/2)
=3
A=3/2