Question

已知函数f(x)=2√3sin(x+π/4)cos(x+π/4)-2sin(x+π)sin

已知函数f(x)=2√3sin(x+π/4)cos(x+π/4)-2sin(x+π)sin(x+5π/2)
求f(x)的最小正周期和单调递增区间
若将f(x)的图像向右平移π/12个单位得到函数g(x)的图像，求g(x)在区间[0,π/2]上的最大值和最小值
要过程，好的我给分，谢谢了

Answer 1

f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3sin(x+π/2)+sinx
=√3cosx+sinx
=2(√3/2cosx+1/2*sinx)
=2sin(x+π/3)
∴f(x)的最小正周期T=2π
(2)将f(x)的图像向右平移π/6个单位,得到函数
g(x)=2sin(x-π/6+π/3)=2sin(x+π/6)
∵x∈[0,π],∴x+π/6∈[π/6,7π/6]
∴x+π/6=π/2时，g(x)取得最大值2
x=7π/6时，g(x)取得最小值-1
请采纳。