问题
填空题
已知f(x)=2sin(x-
|
答案
由f(x)=2sin(x-
)•cos(x-π 4
)+sin2x得π 4
f(x)=sin(2x-
)+sin2x=sin2x-cos2x=π 2
(2
sin2x-2 2
cos2x)=2 2
sin(2x-2
)π 4
根据最小正周期的公式可得:T=
=π2π 2
故答案为π
已知f(x)=2sin(x-
|
由f(x)=2sin(x-
)•cos(x-π 4
)+sin2x得π 4
f(x)=sin(2x-
)+sin2x=sin2x-cos2x=π 2
(2
sin2x-2 2
cos2x)=2 2
sin(2x-2
)π 4
根据最小正周期的公式可得:T=
=π2π 2
故答案为π