Question

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

Answer 1

解：
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
希望对你有帮助，望采纳，谢谢~

Answer 2

1)f(x)=cos(2x+π/3)+sin方x=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2cos2x-根3/2sin2x+1/2-1/2cos2x=1/2-根3/2sin2x
所以最大值为1/2+根3/2，最小正周期为π
2）f（C/2)=-1/4，即1/2-根3/2sinC=-1/4，所以sinC=根3/2，因为C为锐角，所以C=π/6
所以sinA=sin(π-π/6-B)=sin(5π/6-B)=sin5π/6cosB-cos5π/6sinB=(1/2)*(1/3)-(根3/2)*(2根2)/3
=(1-2根6)/6

Answer 3

f(x)=cos(2x+π/3)+sin方x=cox2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2-根3/2*sin2x
因为-1<=sin2x<=1
所以：f(x)=1/2-根3/2*sin2x
1/2-根3/2《=f(x)<=1/2+根3/2
最大值为1/2+根3/2

1/2-根3/2*sinC=-1/4
sinC=根3/2
C=π/3
cosB=1/3
B=arccos1/3
A=π-π/3-arccos1/2
sinA=sin(π/3+arccos1/2)
=sinπ/3cos(arccos1/3)+sin(arccos1/3)cosπ/3
=根3/2*1/3+2根2/3*1/2
=（根3+2根2）/6

Answer 4

f(x)=[(1/2)cos2x－(√3/2)sin2x]＋(1/2)[1－cos2x]=－(√3/2)sin2x＋(1/2)
1、最大值是√3＋(1/2)，最小正周期是π；
2、f(C/2)=－1/4，得：sinC=√3/2，则C=π/3，又cosB=1/3，则sinB=2√2/3
sinA=sin(2π/3－B)=(√3/2)cosB＋(1/2)sinB=[√3＋2√2]/6