Réponses

2014-10-02T11:30:04+02:00
Sin(2π/3+x)=sin(π-π/3+x)=-sin(-π/3+x)=sin(π/3-x)=sin(π/3)cosx-cos(π/3)sinx

sin(4π/3+x)=sin(π+π/3+x)=-sin(π/3+x)=-sin(π/3)cosx-cos(π/3)sinx

Donc
sinx+sin(2π/3+x)+sin(4π/3+x)=sinx+sinxcos(π/3)-cos(π/3)sinx-sinxcos(π/3)-cos(π/3)sinx
sinx+sin(2π/3+x)+sin(4π/3+x))sinx-1/2*sinx-1/2*sinx=sinx-sinx=0