已知向量a=[√2sin(x/2-π/4),√3cosx/2]向量b=[√2sin(x/2+π/4),2sinx/2]
问题描述:
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)
答
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(...