设函数f(x)=cos(2x+π/3)+1/2-1/2cos2x设A,B,C为△ABC的三个内角,若cosB=1/3,f（C/2）=-1/4,且C为锐角,求sinA.

设函数f(x)=cos(2x+π/3)+1/2-1/2cos2x
设A,B,C为△ABC的三个内角,若cosB=1/3,f（C/2）=-1/4,且C为锐角,求sinA.

f(x)=cos(2x+π/3)+1/2-1/2cos2x
f(C/2)=cos(C+π/3)+1/2-cosC/2=-1/4,
cosCcosπ/3-sinCsinπ/3-cosC/2=-3/4,
cosC/2-sinCsinπ/3-cosC/2=-3/4,
-√3/2sinC=-3/4,
sinC=√3/2,cosC=±1/2,
cosB=1/3,sinB=2√2/3,
sinA=sin(B+C)=sinBcosC+cosBsinC
=√3/6±√2/3.