Question

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

Answer 1

解：∵f(x)= cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+cos2xcos(π/2)=cos(2x-π/3) cos(2x-π/3)最小正周期为π ∴ f(x)最小正周期即为π

Answer 2

因为2sin(x-π/4)sin(x+π/4)=cos(x-π/4-x-π/4)-cos(x-π/4+x+π/4)=-cos2x
所以f(x)=cos(2x-π/3)-cos2x=cos(2x-π/6-π/6)-cos(2x-π/6+π/6)=2sin(2x-π/6)sin(π/6)=sin(2x-π/6)
最小正周期是2π/2=π

Answer 3

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x+π/4-π/2)sin(x+π/4)=cos(2x-π/3)-2sin(x+π/4)cos(x+π/4)=cos(2x-π/3)-sin(2x+π/2)=cos(2x-π/3)-cos2x=sin(2x-π/6),T=π

Answer 4

化简之后是Sin(2x-兀/3）所以周期T=兀

Answer 5

T=π