已知a向量=(cos(2x-π/3),sin(x-π/4),b向量=(1,2sin(x+π/4)),函数f(x)=a向量×b向量 （1）求f(x)的对称抽方程（2）求f(x)在区间[-π/12,π/2]上的值域.

Question

已知a向量=(cos(2x-π/3),sin(x-π/4),b向量=(1,2sin(x+π/4)),函数f(x)=a向量×b向量 （1）求f(x)的对称抽方程（2）求f(x)在区间[-π/12,π/2]上的值域.

Answer 1

（1）∵f（x）=a·b
∴f（x）=cos（2x-π/3）×1+sin(x-π/4)×2sin(x+π/4)
=1/2 cos2x+√3/2sin2x+2[﹙√2/2sinx﹚²-﹙√2/2cosx﹚²]
=1/2 cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=sin（2x-π/6）
∴2x-π/6=π/2+kπ,k∈z
解得x=π/3+kπ/2,k∈z
∴对称轴为x=π/3+kπ/2,k∈z
（2）∵-π/12≤x≤π/2
∴-π/6≤2x≤π,
-π/3≤2x-π/6≤5π/6
-√3/2≤sin（2x-π/6）≤1,
∴f(x)在区间[-π/12,π/2]上的值域为[-√3/2,1]