题目应是 若sin(π/6-a)=1/3 则cos(2π/3+3a)=?∵sin(π/6-a)=1/3∴cos[2(π/6-a)]=1-2sin²(π/6-a)=1-2/9=7/9即cos(π/3-2a)=7/9又（π/3-2a）+（2π/3+2a）=π∴cos(2π/3+2a)=cos[π-(π/3-2a)]=-cos(π/3-2...