求函数f(x)=sin(2x+π/6)+cos(2x-π/6)的最小值及相应x的集合不要跳
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/26 10:45:34
求函数f(x)=sin(2x+π/6)+cos(2x-π/6)的最小值及相应x的集合不要跳
求函数f(x)=sin(2x+π/6)+cos(2x-π/6)的最小值及相应x的集合
不要跳
求函数f(x)=sin(2x+π/6)+cos(2x-π/6)的最小值及相应x的集合不要跳
f(x)=sin(2x+π/6)+cos(2x-π/6)
=sin2xcosπ/6+cos2xsinπ/6+cos2xcosπ/6+sin2xsinπ/6
=(√3/2)sin2x+(1/2)cos2x+(√3/2)cos2x+(1/2)sin2x
=[(√3+1)/2](sin2x+cos2x)
=[(√3+1)√2/2](√2/2)(sin2x+cos2x)
=[(√3+1)√2/2]sin(2x+π/4)
∴f(x)最小值就是-(√3+1)√2/2
此时2x+π/4=3π/2+2kπ
x=kπ+5π/8,k属于Z
1L解错了,把正负号弄错了
sin2xcosπ/6 + sinπ/6 cos2x + sin（π/3-2x)
=sin2xcosπ/6 + sinπ/6 cos2x + sinπ/3 cos2x -cosπ/3 sin2x
=(√3/2 -1/2)sin2x + (√3/2+ 1/2)cos2x
=√2sin(2x + β）
tanβ=2+√3
最小值=-√2
f(x)=sin(2x+π/6)+cos(2x-π/6)
=√2[(√2/2)sin(2x+π/6)+√2/2cos(2x-π/6)]
=√2[cos(π/4)sin(2x+π/6)+sin(π/4)cos(2x-π/6)]
=√2sin(2x+π/6+π/4)
=√2sin(2x+5π/12)
所以；函数的最小值=-√2
相应的：2x+5π/12=...
全部展开
f(x)=sin(2x+π/6)+cos(2x-π/6)
=√2[(√2/2)sin(2x+π/6)+√2/2cos(2x-π/6)]
=√2[cos(π/4)sin(2x+π/6)+sin(π/4)cos(2x-π/6)]
=√2sin(2x+π/6+π/4)
=√2sin(2x+5π/12)
所以；函数的最小值=-√2
相应的：2x+5π/12=2kπ+3π/2
2x=2kπ-13π/12
x=kπ-13π/24
相应x的集合:{x|x=kπ-13π/24；k为整数}
收起
f(x)=sin(2x/3)+cos(2x/3)cox(π/6)+sin(2x/3)sin(π/6) ==(根号3)sin(2x/3+π/6) 最大值根号3，最小值－根号3