Question

已知函数f(x)=cos(2x-派/3)+2sin(x-派/4)sin(x+派/4)(1)求函数f(x)的最小正周期(2

已知函数f(x)=cos(2x-派/3)+2sin(x-派/4)sin(x+派/4)(1)求函数f(x)的最小正周期(2)求函数f(x)在区间[-派/12,派/2]上的值域

Answer 1

解:f(x)=cos(2x-π/3)+2sin（x-π/4)sin（x+π/4)
=cos2xcosπ/3+sin2sinπ/3+2(sinxcosπ/4-cosx-sinπ/4)(sinxcosπ/4-cosx-sinπ/4)
=cos2x/2+√3/2sin2x+sin²x-cos²x
=cos2x/2+√3/2sin2x-cos2x
=-cos2x/2+√3/2sin2x
=cos(2x+2π/3）
∴T=2π/ω=2π/2=π
（2）∵-π/12≤x≤π/2
∴π/2≤2x+2π/3≤5π/3
值域为{y|-1≤y≤1/2}

Answer 2

(1)、f(x)=cos(2x-派/3)+2sin(x-派/4)sin(x+派/4)
=cos(2x-派/3)+)+2sin(x-派/4)sin(x-派/4+派/2)
=cos(2x-派/3)+2sin(x-派/4)cos(x-派/4)
=cos(2x-派/3)+sin(2x-派/2)
=cos(2x-派/3)+cos2x
=1/2cos2x+根号3/2sin2x+cos2x
=3/2cos2x+根号3/2sin2x
=根号3*(根号3/2cos2x+1/2sin2x)
=根号3*cos(2x-派/6)
所以函数f(x)的最小正周期为2派/2=派
(2)、函数f(x)在区间[-派/12,派/2]上的值域为[-3/2，根号3]

Answer 3

∏f(x)=cos(2x-派/3)+2sin(x-派/4)sin(x+派/4)
=cos(2x-∏/3)+2sin(x-∏/4)cos(x-∏/4)
=cos(2x-∏/3)+sin(2x-∏/2)
=1/2cos2x-√3/2sin2x-cos2x
=-1/2cos2x-√3/2sin2x
=sin(2x-π/6)
所以最小正周期T=2∏/2=∏
因为函数f(x)在区间[-派/12,派/2]上
所以2x-π/6在区间[-∏/3,5π/6]
即f(x)的值域在[-√3/2，1]

Answer 4

辛苦做的，望及时采纳！