问题 解答题
已知向量
m
=(2cosx,
3
cosx-sinx),
n
=(sin(x+
π
6
),sinx)
,且满足f(x)=
m
n

(I)求函数y=f(x)的单调递增区间;
(II)设△ABC的内角A满足f(A)=2,且
AB
AC
=
3
,求边BC的最小值.
答案

(I)由题意得f(x)=

m
n
=2cosxsin(x+
π
6
)
+(
3
cosx-sinx)sinx

=2

3
sinxcosx+cos2x-sin2x=
3
sin2x+cos2x

=2sin(2x+

π
6
),

由2kπ-

π
2
2x+
π
6
≤2kπ+
π
2
(k∈Z)得,kπ-
π
3
≤x≤kπ+
π
6

则所求的单调递增区间是[kπ-

π
3
kπ+
π
6
](k∈Z).

(Ⅱ)由f(A)=2得,2sin(2x+

π
6
)=2,即sin(2x+
π
6
)
=1,

∵0<A<π,∴

π
6
<2A+
π
6
13π
6
,即2A+
π
6
=
π
2
,解得A=
π
6

AB
AC
=
3
得,bccosA=
3
,解得bc=2,

在△ABC中,a2=b2+c2-2bccosA

=b2+c2-

3
bc≥2bc-
3
bc=(2-
3
)bc
,当且仅当b=c时取等号,

amin2=(2-

3
)×2=4-2
3
,即a=
4-2
3
=
3
-1

单项选择题
问答题 简答题