函数y=2sinxcosx-2sin2x的最小正周期为______.
函数f(x)=2sinxcosx-2sin2x=sin2x+cos2x-1=
sin(2x+2
)+1π 4
所以函数的最小正周期:T=
=π;2π 2
故答案为:π.
函数y=2sinxcosx-2sin2x的最小正周期为______.
函数f(x)=2sinxcosx-2sin2x=sin2x+cos2x-1=
sin(2x+2
)+1π 4
所以函数的最小正周期:T=
=π;2π 2
故答案为:π.
从方框中选词并用其适当形式填空。 | |
2. She doesn't feel well, and she doesn't want to eat ________. 3. It's time ________ sports. 4. Jim ________ a letter to his pen pal once a month last year. 5. I helped my son do some ________ yesterday evening. |