(1)f(30°)=sin90°+2sin(-30°)-cos90°=1-1+0=0, f(60°)=sin120°+2sin0°-cos60°=+0-×=0; (2)由(1)得f(x)=0,证明如下:f(x)=sin(x+60°)+2sin(x-60°)-cos(120°-x) =sinxcos60°+cosxsin60°+2(sinxcos60°-cosxsin60°)-(cos120°cosx+sin120°sinx) =sinx+cosx+2(sinx-cosx)-(-cosx+sinx) =sinx+cosx+sinx-cosx+cosx-sinx)=0 即f(x)=0. |