已知向量m=(根号3sin2x-1,cosx),向量n=(1,2cosx),设函数f(x)=向量m*向量n(1)求f(x)的最小正周期和值域(2)在三角形ABC中,角A、B、C、所对的边分别是a、b、c,若f(A/2)=2,且a^2=bc,试判断三角形ABC的形状

(1)
f(x)
=m.n
=(√3sin2x-1,cosx).(1,2cosx)
=√3 sin2x -1 + 2(cosx)^2
=√3 sin2x -1 + cos2x+1
=2( (√3/2) sin2x + (1/2)cos2x )
=2sin(2x+π/6)
最小正周期 = π
值域 = [-2,2]
(2)
f(A/2) =2sin(A+π/6)= 2
=>sin(A+π/6)= 1
A+π/6 = π/2
A = π/3
by cosine rule
a^2 = b^2 + c^2 - 2bccosA
= b^2 + c^2 - 2(a^2)(1/2)
a^2 = (b^2+c^2)/2

f(x) =m*n=(√3sin2x 2,cosx)*(1,2cosx) =√3sin2x 2 2(cosx)^2 =√3sin2x 2*(1 cos2x)/2 2 =√3sin2x cos2x) 3 =2sin(2x ∏/6)

(1)f(x)=[√3sin(2x)-1]*1+cosx*2cosx=√3sin(2x)-1+2(cosx)^2=√3sin(2x)+cos(2x)=2sin(2x+π/6)所以,f(x)的最小正周期T=2π/2=π,值域是[-2,2].(2)f(A/2)=2sin(2*A/2+π/6)=2,得A=π/3由余弦公式知a^2=b^2+c^2-2bc...