已知向量a=[√2sin(x/2-π/4),√3cosx/2]向量b=[√2sin(x/2+π/4),2sinx/2]函数f（x）=向量a·向量b1.求函数f（x）的对称轴方程及其单调递增区间2.在锐角△ABC中,若f（A）=2/3,求cosA的值（这题可答可不答）

Question

已知向量a=[√2sin(x/2-π/4),√3cosx/2]向量b=[√2sin(x/2+π/4),2sinx/2]
函数f（x）=向量a·向量b
1.求函数f（x）的对称轴方程及其单调递增区间
2.在锐角△ABC中,若f（A）=2/3,求cosA的值（这题可答可不答）

Answer 1

f（x）=a·b=√2sin(x/2-π/4)√2sin(x/2+π/4)+2√3cosx/2sinx/2
=2sin(x/2-π/4)sin[π/2-(π/4-x/2)]+2√3cosx/2sinx/2
=-2sin(x/2-π/4)cos(x/2-π/4)+√3sinx
=-sin(x-π/2)+√3sinx
=cosx+√3sinx
=2sin(x+π/6)
f（x）的对称轴:x+π/6=kπ
x=kπ-π/6
x+π/6在[kπ-π/2,2kπ+π/2]单调递增
x在[2kπ-2π/3,2kπ+π/3]单调递增
f（A）=2sin(x+π/6)=2/3
sin(x+π/6)=1/3
cos(x+π/6)=2√2/3
cosA=cosA[(x+π/6)-π/6]=√3/2cosA(x+π/6)+1/2sin(x+π/6)
=√3/2*2√2/3+1/2*1/3
=(1+√6)/6