已知a属于（0,π）,cos(π/3+a)=1/3,求sin（π/3-2a),

cos(π/3+a)=1/3
→cos^2 (π/3+a)=1/9
→2cos^2 (π/3+a)-1=-7/9;
即cos(2π/3+2a)=-7/9;
则cos(-2π/3-2a)=-7/9;
则cos[(-2π/3-2a)+π]=+7/9;
即cos（π/3-2a)=7/9;
a∈（0,π）,则
π/3+a∈（π/3,4π/3）;
而cos(π/3+a)=1/3＞0,
则说明π/3＜π/3+a＜π/2．
→0＜a＜π/6;
→π/3-2a∈(0,π/3);
则sin（π/3-2a)>0;
sin（π/3-2a)=√[1-cos^2 (π/3+a)]
=(2√2)/3