文科数学三角函数 1.已知：a=(2cosx,sinx),b=(√3cosx,2cosx).设函数f(x)-axb-√3

1（1）f(x)=a●b-√3
=2√3cos²x+2sinxcosx-√3
= √3(1+cos2x)+sin2x-√3
=sin2x+√3cos2x
=2(1/2*sin2x+√3/2*cos2x)
=2sin(2x+π/3)
f(x)的最小正周期;T=π
(2)
由2kπ-π/2≤2x+π/3≤2kπ+π/2,k∈Z
得:kπ-5π/12≤x≤kπ+π/12,k∈Z
∴f（x）的单调递增区间是
[kπ-5π/12,kπ+π/12],k∈Z
f（x）的单调递减区间是
[kπ+π/12,kπ+7π/12],k∈Z
（3）
∵x∈[-π/4,π/4∴2x+π/3∈[-π/6,5π/6]
∴2x+π/3=π/2,f(x)max=2
2x+π/3=π/2,f(x)min=-1
∴f(x)的值域为[-1,2]
2
f(x)=sin(2x-[π/4])-2√2(sin^2)x
=√2/2*sin2x-√2/2*cos2x-√2(1-cos2x)
=√2/2sin2x+√2/2*cos2x-√2
=sin(2x+π/4)-√2
∵x∈R∴-1≤sin(2x+π/4)≤1
∴2x+π/4=-π/2+2kπ,k∈Z时,
f(x)取得最小值-1-√2