(2014•四川二模)已知函数f(x)=23sinxcosx-3sin2x-cos2x+3.
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/11/08 01:32:42
(2014•四川二模)已知函数f(x)=2
3 |
(1)∵f(x)=2
3sinxcosx-3sin2x-cos2x+3
=
3sin2x-3•
1−cos2x
2-
1+cos2x
2+3
=
3sin2x-cos2x+1=2sin(2x+
π
6)+1,
∵x∈[0,
π
2],∴2x+
π
6∈[
π
6,
7π
6],
∴sin(2x+
π
6)∈[−
1
2,1],
∴f(x)=2sin(2x+
π
6)+1∈[0,3];
(2)∵
sin(2A+C)
sinA=2+2cos(A+C),
∴sin(2A+C)=2sinA+2sinAcos(A+C),
∴sinAcos(A+C)+cosAsin(A+C)=2sinA+2sinAcos(A+C),
∴-sinAcos(A+C)+cosAsin(A+C)=2sinA,即sinC=2sinA,
由正弦定理可得c=2a,又由
b
a=
3可得b=
3sinxcosx-3sin2x-cos2x+3
=
3sin2x-3•
1−cos2x
2-
1+cos2x
2+3
=
3sin2x-cos2x+1=2sin(2x+
π
6)+1,
∵x∈[0,
π
2],∴2x+
π
6∈[
π
6,
7π
6],
∴sin(2x+
π
6)∈[−
1
2,1],
∴f(x)=2sin(2x+
π
6)+1∈[0,3];
(2)∵
sin(2A+C)
sinA=2+2cos(A+C),
∴sin(2A+C)=2sinA+2sinAcos(A+C),
∴sinAcos(A+C)+cosAsin(A+C)=2sinA+2sinAcos(A+C),
∴-sinAcos(A+C)+cosAsin(A+C)=2sinA,即sinC=2sinA,
由正弦定理可得c=2a,又由
b
a=
3可得b=
(2014•四川二模)已知函数f(x)=23sinxcosx-3sin2x-cos2x+3.
(2014•四川二模)已知函数f(x)=cos2x-sin2x+23sinxcosx.
(2014•甘肃二模)已知函数f(x)=sin2x+23sinxcosx+3cos2x.
(2009•台州二模)已知函数f(x)=sin2x+23sinxcosx+3cos2x.
(2013•红桥区二模)已知函数f(x)=sin2x+2sinxcosx+3cos2x.
(2013•潮州二模)已知函数f(x)=3(sin2x−cos2x)−2sinxcosx.
(2014•马鞍山一模)已知函数f(x)=23sinxcosx−3sin2x−cos2x+2.
(2014•大港区二模)已知函数f(x)=sinxcosx+3cos2x.
(2014•东城区二模)已知函数f(x)=sin2x+3sinxcosx.
(2014•达州二模)设函数f(x)=53sinxcosx+6cos2x+sin2x+32.
(2012•南昌模拟)已知函数f(x)=3sin2x+23sinxcosx+5cos2x. &nb
(求问)已知函数f(x)=sin2x+2根号3sinxcosx-cos2x