Question

已知向量a=(2cosx,1),b=(cosx,根号3sin2x-1),设函数f(x)=向量a*向量b，其中x∈R（1）求函数的最小正周期

和单调递增区间

Answer 1

解答：
f(x)=a.b
=(2cosx,1).(cosx,√3sin2x-1)
=2cos²x+√3sin2x-1
=√3sin2x+cos2x
=2[sin(2x)*(√3/2)+cos(2x)*(1/2)]
=2[sin(2x)cos(π/6)+cos(2x)sin(π/6)]
=2sin(2x+π/6)
(1) T=2π/2=π
(2) 2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
所以，增区间为【kπ-π/3, kπ+π/6】,k∈Z

Answer 2

f(x)=ab=2cos²x+√3sin2x-1
=2cos²x-1+√3sin2x
=cos2x+√3sin2x
=2[(1/2)cos2x+(√3/2)sin2x]=2(sinπ/6cos2x+cosπ/6sin2x)
=2sin(2x+π/6)
(1).T=2π/2=π
(2)-π/2+2kπ≤2x+π/6≤π/2+2kπ
-2π/3+2kπ≤2x≤π/3+2kπ
-π/3+kπ≤x≤π/6+kπ k=0,1,2....
单调递增区间[-π/3+kπ,π/6+kπ] k=0,1,2....

追问

若将函数f（x）的图像的纵坐标保持不变，横坐标扩大到原来的的两倍，然后再向右平移π/6个单位得到g（x）的图像，求其解析式

追答

g(x)=2sin[(1/2)*2(x-π/6)+π/6]=2sinx