(2011•顺义区二模)已知函数f(x)=2−sin(2x+π6)−2sin2x,x∈R
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:综合作业 时间:2024/07/22 06:34:44
(2011•顺义区二模)已知函数f(x)=2−sin(2x+
)−2sin
π |
6 |
![(2011•顺义区二模)已知函数f(x)=2−sin(2x+π6)−2sin2x,x∈R](/uploads/image/z/386026-34-6.jpg?t=%EF%BC%882011%E2%80%A2%E9%A1%BA%E4%B9%89%E5%8C%BA%E4%BA%8C%E6%A8%A1%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9D2%E2%88%92sin%282x%2B%CF%806%29%E2%88%922sin2x%EF%BC%8Cx%E2%88%88R)
(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=
(2011•顺义区二模)已知函数f(x)=2−sin(2x+π6)−2sin2x,x∈R
(2011•顺义区一模)已知函数f(x)=2sinxcosx+23sin2x−3,(x∈R).
(2011•朝阳区二模)已知函数f(x)=2sinx•sin(π2+x)−2sin2x+1(x∈R).
(2014•烟台一模)已知函数f(x)=sin(7π6−2x)−2sin2x+1(x∈R),
(2012•资阳一模)已知函数f(x)=[2sin(x−π3)+sinx]•cosx+3sin2x(x∈R).
(2011•武汉模拟)已知函数f(x)=[2sin(x+π3)+sinx]cosx−3sin2x,x∈R
(2013•资阳模拟)已知函数f(x)=2sin(x−π3)cosx+sinxcosx+3sin2x(x∈R).
已知函数f(x)=2sin^2x+sin2x-1,x∈R
(2014•东莞二模)已知函数f(x)=2sin2x+2cos2x,x∈R.
(2011•深圳二模)设函数f(x)=sinωx+sin(ωx−π2),x∈R.
已知函数f(x)=2sin(x−π4)•sin(x+π4)+sin2x
已知函数f(x)=2sin(13x−π6),x∈R