Question

设函数f（x）=cos（2x+π/3）+sin²x，求函数的最小正周期

要过程

Answer 1

f(x)=cos(2x+π/3)+(sinx)^2
=(cos2x)/2+√3(sin2x)/2+(1-cos2x)/2 (∵竖码老cos2x=1-2(sinx)^2,∴(sinx)^2=(1-cos2x)/2)
=1/2+√3(sin2x)/2
∴函余升数最模则小正周期为π

Answer 2

f(x) = cos(2x+π/3) + sin²x
= cos(2x+π/3) + (1 - cos2x) / 2
= cos(2x+π/3) - cos2x / 2 + 1/2
= [2cos(2x+π/3) - cos2x] / 2 + 1/2
= [2(cos2x * cosπ/3 - sin2x * sinπ/3) - cos2x] / 2 + 1/2
= [cos2x - √3sin2x - cos2x] / 2 + 1/2 ----cosπ/岁仿3=1/2；sinπ/3=√3/2
= - √3sin2x / 2 + 1/2
函数的最小正周期带雀毕为： 2π/2 = π
补充：蠢芹最大值为 (1 + √3) / 2
最小值为 (1 - √3) / 2

Answer 3

cos（2x+π/3）+sin²x=1/2*(cosx的平方-siny的平方)+根号下喊哪3/2sin2x+sinx的平方=1/2(cosx的此渗激森袜平方+siny的平方)+根号下3/2sin2x=1/2+根号下3/2sin2x,则最小正周期为怕（不好意思那个符号没找着）

Answer 4

f(x) = cos(2x+π/3) + sin²x
= cos(2x+π/3) + (1 - cos2x) / 2
= cos(2x+π/3) - cos2x / 2 + 1/2
= [2cos(2x+π/3) - cos2x] /搏余竖 2 + 1/2
= [2(cos2x * cosπ/3 - sin2x * sinπ/3) - cos2x] / 2 + 1/2
= [cos2x - √3sin2x - cos2x] / 2 + 1/2 ----cosπ/3=1/2；sinπ/3=√3/2
= - √3sin2x / 2 + 1/2
函数的最小正周期为： 2π/2 = π
补充：最大值为 (1 + √基大3) / 2
最小值毁悔为 (1 - √3) / 2