暗夜多思
幼苗
共回答了22个问题采纳率:81.8% 举报
(I)∵ f(x)=cos(2x+
π
3 )+si n 2 x
=
1
2 cos2x-
3
2 sin2x+
1-cos2x
2
=
1
2 -
3
2 sin2x
又 0≤x≤
π
6 ∴ 0≤2x≤
π
3 0≤sin2x≤
3
2
则 -
1
4 ≤f(x)≤
1
2
函数f(x)的值域[ -
1
4 ,
1
2 ]
(II)∵cosB=
1
3 ∴sinB=
2
2
3
∵ f(
C
2 ) =
1
2 -
3
2 sinC=-
1
4
∴ sinC=
3
2 且0°<C<90° 则C=60°
∴sinA=sin(120°-B)=
3
2 cosB+
1
2 sinB =
3
2 ×
1
3 +
1
2 ×
2
2
3 =
3 + 2
2
6
1年前
7