(2014•马鞍山一模)已知函数f(x)=23sinxcosx−3sin2x−cos2x+2.
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(2014•马鞍山一模)已知函数f(x)=2
sinxcosx−3sin
3 |
(Ⅰ)∵f(x)=
3sin2x-3sin2x-cos2x+2(sin2x+cos2x)
=
3sin2x+cos2x-sin2x
=
3sin2x+cos2x
=2sin(2x+
π
6),
∴f(x)的最大值是2;
(Ⅱ)由条件得sin(2A+C)=2sinA+2sinAcos(A+C),
即sinAcos(A+C)+cosAsin(A+C)=2sinA+2sinAcos(A+C),
化简得sinC=2sinA,由正弦定理得:c=2a,
又b=
3a,
由余弦定理得:a2=b2+c2-2bccosA
=3a2+4a2-4
3a2cosA,
整理得:cosA=
3
2,A∈(0,π),
∴A=
π
6,B=
π
3,C=
π
2,
∴f(B)=f(
π
3)=2sin
5π
6=1.
3sin2x-3sin2x-cos2x+2(sin2x+cos2x)
=
3sin2x+cos2x-sin2x
=
3sin2x+cos2x
=2sin(2x+
π
6),
∴f(x)的最大值是2;
(Ⅱ)由条件得sin(2A+C)=2sinA+2sinAcos(A+C),
即sinAcos(A+C)+cosAsin(A+C)=2sinA+2sinAcos(A+C),
化简得sinC=2sinA,由正弦定理得:c=2a,
又b=
3a,
由余弦定理得:a2=b2+c2-2bccosA
=3a2+4a2-4
3a2cosA,
整理得:cosA=
3
2,A∈(0,π),
∴A=
π
6,B=
π
3,C=
π
2,
∴f(B)=f(
π
3)=2sin
5π
6=1.
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