化简f(x)=cos[(6k+1)π/3+2x]+cos[(6k-1)π/3-2x]+2√3sin(π/6-2x) x∈R,k∈Z
网友回答
cos[(6k+1)π/3+2x]=cos[2kπ+π/3+2x]=cos[π/3+2x]
cos[(6k-1)π/3-2x]=cos[2kπ-π/3-2x]=cos[π/3+2x]
那么原式=2cos[π/3+2x]+2√3sin(π/6-2x)
2√3sin(π/6-2x)=2√3sin[π/2-(π/3+2x)]=2√3cos(π/3+2x)
原式=（2+2√3）cos(π/3+2x)