Question

设函数f(x)=(sinwx+coswx)平方+2cos平方wx(w>0)的最小正周期为2派/3。

1,求W的值2,若函数y=g(x)的图像是由y=f(x)的图像向右平移派/2个单位长度得到,求y=g(x)的单调增区间

Answer 1

f(x)=(sinwx+coswx)^2+2(coswx)^2
=1+2sinwxcoswx+2(coswx)^2
=sin2wx+cos2wx+2
=√2sin(2wx+π/4)+2
1,最小正周期T=2π/2w=2π/3，则w=3/2。
2,f(x)=√2sin(3x+π/4)+2向右平移π/2个单位。
则g(x)=√2sin[3(x-π/2)+π/4]+2=√2sin(3x-5π/4)+2=√2sin(3x-π/亏春携4)+2。
2kπ-π/2<=3x-π/4<=2kπ+π/2，则2kπ/销伏3-π/12<=x<=2kπ/3+π/4。
g(x)的单调增长率区间是[2kπ/3-π/12,2kπ/3+π/4]，k为整森渣数。

Answer 2

f(x)=(sinwx+coswx)^2+2(coswx)^2
=1+2sinwxcoswx+2(coswx)^2
=sin2wx+cos2wx+2
=√2sin(2wx+π/4)+2
最小正周启派期T=2π/2w=2π/亏旁族3，则w=3/2。
2:f(x)=√2sin(3x+π/4)+2向右平移π/销弊2个单位
则有:g(x)=√2sin[3(x-π/6)+π/4]
=√2sin(3x-π/4)
令2kπ-π/2<=3x-π/4<=2kπ+π/2,k为整数
所以g(x)的单增区间为[2kπ/3-π/12,2kπ/3-3π/12],k为整数