(1) a.b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=cos(2x)(2) |a+b|=√{[cos(3x/2)+cos(x/2)]^2+[sin(3x/2)-sin(x/2)]^2}=√[2+2cos(3x/2)cos(x/2)-2sin(3x/2)sin(x/2)]=√[2+2cos(2x)]=2cos(x)(3) f(x)=cos(2x)-2cos(x...为什么=√[2+2cos(2x)]=2cos(x)用倍角公式 cos(2x)=2[cos(x)]^2-1,代入即得。