数学三角函数,a=( 2sin(x+π/12),cos(x-π/12) ) b=( cos(x+π/12),2sin(x-π/12) ),f(x)=a*b-2cos²x 1.求f(x)的最小正周期 2.y=g(x)的图像向左平移π/4个单位长度,在向下平移1个单位长度得到,当x∈(0,π/2)时,求y=g(x)的最大值和最小值
问题描述：
数学三角函数,
a=( 2sin(x+π/12),cos(x-π/12) ) b=( cos(x+π/12),2sin(x-π/12) ),f(x)=a*b-2cos²x 1.求f(x)的最小正周期 2.y=g(x)的图像向左平移π/4个单位长度,在向下平移1个单位长度得到,当x∈(0,π/2)时,求y=g(x)的最大值和最小值
答
f(x)=向量a.向量b-2cos^2x.
=2sin（x+π/12)cos(x+π/12)+2sin(x-π/12)cos(x-π/12)-2cos^2x.
=sin(2x+π/6)+sin(2x-π/6)-2cos^2x.
=2sin[(2x+π/6+2x-π/6)/2]cos[(2x+π/6-2x+π/6)]/2-2cos^2x.
=2sin2xcosπ/6-(1+cos2x).
=2[(√3/2)sin2x-(1/2)cos2x-1.
∴f(x)=2sin(2x-π/6)-1.
1. f(x)的最小正周期T=2π/2=π.
2.y=g(x).
g(x)与fx)有什么关系?