Question

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

Answer 1

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