题目内容：

设函数f(x)=cos(2x+π/3)+sin方x
1）求函数f(x)的最大值和最小正周期
2）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

优质解答

f(x)=cos(2x+π/3)+sin²x
=cos2xcosπ/3-sin2xsinπ/3+[1-cos(2x)]/2
=1/2cos2x-√3/2sin2x+1/2-1/2cos2x
=-√3/2sin2x+1/2
当sin2x=-1时,最大值=(1+√3)/2
最小正周期=2π/2=π
f(C/2)=-√3/2sinC+1/2=-1/4
sinC=√3/2
∵C为锐角
C=π/3
cosC=1/2
∵cosB=1/3
∴sinB=2√2/3
sinA=sin(π-B-C)
=sin(B+C)
=sinBcosC+cosBsinC
=2√2/3×1/2+1/3×√3/2
=√2/3+√3/6