已知函数f(x)=2√3sinx(x+π/4)cos(x+π/4)-sin(2x+π)
问题描述：
已知函数f(x)=2√3sinx(x+π/4)cos(x+π/4)-sin(2x+π)
答
f(x)=2√3sinx(x+π/4)cos(x+π/4)-sin(2x+π)
=√3*2sinx(x+π/4)cos(x+π/4)-sin(2x+π)
=√3sin(2x+π/2)+sin2x
=√3cos2x+sin2x
=2(1/2*sin2x+√3/2*cos2x)
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)
手机提问的朋友在客户端右上角评价点【满意】即可求函数的最小正周期和单调递增区间 若将fx的图像向右平移π/12个单位，得到函数gx的图像求函数gx在区间[0，π/2]上的最大值和最小值最小正周期=2π/2=π令-π/2+2kπ