(2011•顺义区二模)已知函数f(x)=2−sin(2x+π6)−2sin2x,x∈R
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(2011•顺义区二模)已知函数f(x)=2−sin(2x+
)−2sin
π |
6 |
(1)f(x)=2−sin(2x+
π
6)−2sin2x
=2−(sin2xcos
π
6+cos2xsin
π
6)−(1−cos2x)
=1+cos2x−(
3
2sin2x+
1
2cos2x)
=
1
2cos2x−
3
2sin2x+1
=cos(2x+
π
3)+1…(5分)
∵ω=2,∴T=
2π
2=π,
则函数f(x)的最小正周期为π;
(2)由f(
B
2)=1得:cos(B+
π
3)+1=1,即cos(B+
π
3)=0,
又0<B<π,∴
π
3<B+
π
3<
4
3π,
∴B+
π
3=
π
2,即B=
π
6,…(9分)
∵b=1,c=
3,
∴由正弦定理
b
sinB=
c
sinC得:sinC=
π
6)−2sin2x
=2−(sin2xcos
π
6+cos2xsin
π
6)−(1−cos2x)
=1+cos2x−(
3
2sin2x+
1
2cos2x)
=
1
2cos2x−
3
2sin2x+1
=cos(2x+
π
3)+1…(5分)
∵ω=2,∴T=
2π
2=π,
则函数f(x)的最小正周期为π;
(2)由f(
B
2)=1得:cos(B+
π
3)+1=1,即cos(B+
π
3)=0,
又0<B<π,∴
π
3<B+
π
3<
4
3π,
∴B+
π
3=
π
2,即B=
π
6,…(9分)
∵b=1,c=
3,
∴由正弦定理
b
sinB=
c
sinC得:sinC=
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