Question

已知函数f(x)=cos（2X-π/3）+2sin（X-π/4）cos（X-π/4）（X∈R），求函数F（X）最小正周期。

求函数F（X）在区间【-π/12，π/2】上的值域.

Answer 1

f(x)=cos（2X-π/3）+2sin（X-π/4）cos（X-π/4）（X∈R），
=cos(2x-π/3)+sin(2x-π/2)=sin(2x-π/3+π/2)+sin(2x-π/2)=sin(2x+π/6)+sin(2x-π/2)
=2sin[(2x+π/6+2x-π/2)/2]cos[(2x+π/6-2x+π/2)/2]=2sin(2x-π/6)cos(π/3)=sin(2x-π/6)
其最小正周期为2π/2=π
x∈[-π/12，π/2]
2x-π/6∈[-π/3，5π/6]
在区间【-π/12，π/2】上的值域为[-√3/2，1]

Answer 2

cos（2X-π/3）=cos2Xcosπ/3+sin2Xsinπ/3=1/2*cos2X+√3/2*sin2X
2sin（X-π/4）cos（X-π/4）=sin（2X-π/2）=-cos2X
f(x)=-1/2*cos2X+√3/2*sin2X=cos2Xcos2π/3+sin2Xsin2π/3=cos(2X-2π/3)
f(x)的最小正周期T=2π/w=π
x属于[-π/12，π/2],则2X-2π/3属于[-5π/6,π/3],函数f(x)在区间[-π/12，π/2]上的值域.:[-√3/2,1]