设函数f(x)=向量a.b,其中向量a=(2cosx,1),b=(cosx,√3sin2x)x属于
1个回答

1.f(x)=a*b=2cos²x+√3sin2x

=cos2x+√3sin2x+1

=2sin(2x+Π/6)+1

最小正周期T=2Π/2=Π

2.f(A)=2sin(2A+Π/6)+1=2

2sin(2A+Π/6)=1

sin(2A+Π/6)=1/2

则2A+Π/6=Π/6或者5Π/6

A≠0,所以A=Π/3

由正弦定理知：a/sinA=b/sinB=c/sinC=(b+c)/(sinB+sinC)

得sinB+sinC=3/2

即2sin[(B+C)/2]cos[(B-C)]=3/2

cos(B-C)=√3/2

B-C=Π/6

B+C=2Π/3

得出B=5Π/12 C=Π/4

由正弦定理知：a/sinA=c/sinC

得：c=√2

则b=3-√2

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