Question

求函数y=2cos（x+π/4）cos（x-π/4）+√3sin2x在｛π/4，5π/6｝的值域和最小正周期

Answer 1

y=2*(1/2)[cos(2x)-sin(π/2)]+√3sin2x
=cos2x+√3sin2x-1
=2sin(2x+30°)-1.

所以最小正周期为2π/2=π.
因为：
-1<=sin(2x+30°)<=1
-2<2sin(2x+30°)<=2
-3<2sin(2x+30°)-1<=1

所以值域为。[-3,1].

Answer 2

y=2cos（x+π/4）cos（x-π/4）+√3sin2x
=2cos（x+π/4）sin（x+π/4）+√3sin2x
=sin(2x+π/2)+√3sin2x
=cos2x+√3sin2x
=2sin(2x+π/6)
所以最小正周期T=2π/2=π
因π/4<x<5π/6
所以2π/3<2x+π/6<11π/6
-1<sin(2x+π/6)<√3/2
-2<2sin(2x+π/6)<√3
在｛π/4，5π/6｝的值域为(-2,√3)

Answer 3

y=2cos（x+π/4）cos（x-π/4）+√3sin2x
=（cosx-sinx）（sinx+cosx）+√3sin2x
=（cosx）^2-(sinx)^2+√3sin2x
=cos2x+√3sin2x
=2sin(π/6+2x)
在（π/4，5π/6）的值域为[-2,根号3），最小正周期π