Question

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

在三角形ABC中a，b，c分别是角A，B，C的对边，f（A）=2,A=根号3,b+c=3（b>c），求b，c的长

Answer 1

f(x)=ab
=2cosx*cosx+1*√3sin2x
=cos2x+1+√3sin2x
=cos2x+√3sin2x+1
=2(1/2*cos2x+√3/2*sin2x)
=2(sinπ/6*cos2x+cos2π/6*sin2x)
=2sin(2x+π/6)

T=2π/2=π

Answer 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