由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) 又2√3sin(π/6-2x)=2√3sin[π/2-(π/3+2x)]=2√3cos(π/3+2x...