设函数f(x)=(sinwx+coswx)^2+2cos^2wx(w>0)的最小正周期为2π/3
问题描述:
设函数f(x)=(sinwx+coswx)^2+2cos^2wx(w>0)的最小正周期为2π/3
1求w的值
2若函数y=g(x)的图像由y=f(x)的图像向右平移π/2个单位长度得到,求y=g(x)的单调增区间
答
1.f(x)=(sinwx+coswx)^2+2cos^2wx
=1+sin2wx+1+cos2wx
=2+根号2sin(2wx+π/4)
T=2π/2w=2π/3,w=3/2
2.g(x)=2+根号2sin[3(x-π/2)+π/4]
=2+根号2sin(3x-5π/4)
令-π/2+2Kπ