原式= sin(π/3-2a)+cos(π/3+2a)= sin(π/3-2a)+sin〔π/2-(π/3+2a) 〕= sin(π/3-2a)+sin(π/6-2a) =2sin(π/4-2a)cos(π/12)=2 cos(π/12) sin(π/4-2a)因为cos(π/12)= cos[(π/6)/2]= √{[1+cos(π/6)]/2}=[√...