(1)a·b=cos3x/2cosx/2-sin3x/2sinx/2=cos(3x/2+x/2)=cos2x|a+b| =√[(cos3x/2+cosx/2)^2+(sin3x/2-sinx/2)^2]=√[(cos3x/2)^2+(sin3x/2)^2+(cosx/2)^2+(sinx/2)^2+2cos3x/2cosx/2-2sin3x/2sinx/2]=√(2+2(cos3x/2co...=√(2+2(cosx)^2-2(sinx)^2)=√4(cosx)^2这步是怎么化得根据二倍角公式：cos2x=cos²x-sin²x所以√(2+2cos2x) =√(2+2(cosx)^2-2(sinx)^2)=√(2(cosx)^2+2-2(sinx)^2)=√(2(cosx)^2+2(1-(sinx)^2))=√(2(cosx)^2+2(cosx)^2)=√4(cosx)^2=2cosx