Question

已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4).

(1)求函数f(x)的最小正周期和图象的对称轴方程
(2)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)sin(x+π/4)
=(cos2x)/2+（ 根号3*sin2x）/2+（ sin2x+cos2x ）(sin2x-cos2x)=(-1/2)cos2x+ 根号3*sin2x）/2=cos(2x+π/6)
最小正周期为11π/12，对称轴x=kπ/2-π/12,k 为整数

2.k=0,x=-π/12, k=1,x=5π/12
f(x)在[-π/12,5π/12]上单调递减
当x=-π/12时取到最大值为1,当x=5π/12时取到最小值为-1,值域为[-1,1]