荼啡茶 > 已知函数f(x)=cos(2x-pai⼀3)+2sin(x-pai⼀4).sin(x+pai⼀4)求函数在区间[-pai⼀12,pai ⼀12]上最大值和最小值

已知函数f(x)=cos(2x-pai⼀3)+2sin(x-pai⼀4).sin(x+pai⼀4)求函数在区间[-pai⼀12,pai ⼀12]上最大值和最小值

2025-05-07 02:36:46

推荐回答（2个）

回答1：

f(x)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)
=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cos(2x-π/3)+sin(2x-π/2)
=cos(2x-π/3)-cos2x
=(1/2)cos2x+(√3/2)sin2x-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
﹣π/12≤x≤π/12
﹣π/6≤2x≤π/6
﹣π/3≤2x-π/6≤0
当2x-π/6=0时，y最大
y(max)=0
当2x-π/6= -π/3时，y最小
y(min)= - √3/2

回答2：

楼主确定2sin(x-pai/4).sin(x+pai/4)的符号没错么？