已知函数f(x)=2sin(ωx-π6)sin(ωx+π3)(其中ω为正常数,x∈R)的最小正周期为π.(1)求ω的值;(2)在△ABC中,若A<B,且f(A)=f(B)=12,求BCAB.-数学

题目简介

已知函数f(x)=2sin(ωx-π6)sin(ωx+π3)(其中ω为正常数,x∈R)的最小正周期为π.(1)求ω的值;(2)在△ABC中,若A<B,且f(A)=f(B)=12,求BCAB.-数学

题目详情

已知函数f(x)=2sin(ωx-
π
6
)sin(ωx+
π
3
)
(其中ω为正常数,x∈R)的最小正周期为π.
(1)求ω的值;
(2)在△ABC中,若A<B,且f(A)=f(B)=
1
2
,求
BC
AB
题型:解答题难度:中档来源:深圳一模

答案

(1)∵f(x)=2sin(ωx-class="stub"π
6
)sin(ωx+class="stub"π
3
)=2sin(ωx-class="stub"π
6
)cos[(ωx+class="stub"π
3
)-class="stub"π
2
]

=2sin(ωx-class="stub"π
6
)cos(ωx-class="stub"π
6
)
=sin(2ωx-class="stub"π
3
)
.(4分)
而f(x)的最小正周期为π,ω为正常数,
class="stub"2π
,解之,得ω=1.(6分)
(2)由(1)得f(x)=sin(2x-class="stub"π
3
)

若x是三角形的内角,则0<x<π,
-class="stub"π
3
<2x-class="stub"π
3
<class="stub"5π
3

f(x)=class="stub"1
2
,得sin(2x-class="stub"π
3
)=class="stub"1
2

2x-class="stub"π
3
=class="stub"π
6
2x-class="stub"π
3
=class="stub"5π
6

解之,得x=class="stub"π
4
x=class="stub"7π
12

由已知,A,B是△ABC的内角,A<B且f(A)=f(B)=class="stub"1
2

A=class="stub"π
4
B=class="stub"7π
12
,∴
C=π-A-B=class="stub"π
6
.(10分)
又由正弦定理,得class="stub"BC
AB
=class="stub"sinA
sinC
=
sinclass="stub"π
4
sinclass="stub"π
6
=
2
2
class="stub"1
2
=
2
.(12分)

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