Question

已知f(x)=向量a乘以向量b，其中向量a=(2cosx,-根号3的sin2x),向量b=(cosx,1)(x属于R)

1)求f(x)的周期和递减区间 2)在△ABC中，角A,B,C的对边分别为a,b,c,f(A)=-1，a=√7向量AB*向量AC=3，求边长b和c的值(b>c)

Answer 1

1
f(x)=a·b=(2cosx,-√3sin(2x))·(cosx,1)=2cosx^2-√3sin(2x)
=1+cos(2x)-√3sin(2x)
=1+2cos(2x+π/3)
最小正周期:T=2π/2=π
减区间:2x+π/3∈[2kπ,2kπ+π]
即:x∈[kπ-π/6,kπ+π/3]，k∈Z
2
f(A)=1+2cos(2A+π/3)=-1
即:cos(2A+π/3)=-1
2A+π/3∈(π/3,7π/3)
故:2A+π/3=π，即:A=π/3
AB·AC=|AB|*|AC|*cosA=bccos(π/3)=3
即:bc=6
a^2=b^2+c^2-2bccosA=b^2+c^2-bc=7
即:b^2+c^2=13，即:(b+c)^2=25
即:b+c=5，即:b^2-5b+6=0
即:b=2或3，c=3或2
b>c，故:b=3，c=2

Answer 2

f(x)=向量a乘以向量b
=2cosx*cosx+1*√3sin2x
=cos2x+1+√3sin2x
=2(1/2*cos2x+√3/2*sin2x)
=2sin(2x+π/6)
fx最小正周期=2π/2=π
增区间是
2kπ-π/2<=2x+π/6<=2kπ+π/2
2kπ-2π/3<=2x<=2kπ+π/3
kπ-π/3<=x<=kπ+π/6
∴在［0，π］上的单调递增区间
是[0,π/6]u[π2/3,π]
2
a^2+b^2-c^2>=ab
(a^2+b^2-c^2)/ab>=1
∵cosc=(a^2+b^2-c^2)/2ab
∴2cosc=(a^2+b^2-c^2)/ab>=1
1>cosc>=1/2
0<c<=π/3
f(c)=2sin(2c+π/6)
∵0<c<=π/3
π/6<2c+π/6<=5π/6
∴当2c+π/6=π/2时得最大值2
当2c+π/6=5π/6时得最小值1
所以f(c)的范围是[1,2]