f(x)=cos(2x+π/3)+sin²x=1/2cos2x-sinπ/3sin2x+(1-cos2x)/2=-√3/2*sin2x+1/2
1/2cos2x-sinπ/3sin2x怎么来的?
先看这几个三角基本公式:cos(a+b)=cosa*cosb-sina*sinb(cos(a-b)=cosa*cosb+sina*sinb)sin(a+b)=sina*cosb+cosa*sinb(sin(a-b)=sina*cosb-cosa*sinb)故cos(2x+π/3)=cos2x*cosπ/3-sin2x*sinπ/3=1/2cos2x-sinπ/3sin2x