设函数f(x)定义域为R,周期为π,且f(x)=sinx,-π2≤x<0cosx,0≤x<π2,则f(-5π3)= _ .

问题描述:

设函数f(x)定义域为R,周期为π,且f(x)=

sinx,-
π
2
≤x<0
cosx,0≤x<
π
2
,则f(-
3
)= ___ .

已知:函数f(x)定义域为R,周期为π,
∴f(-

3
)=f(-2π+
π
3
)=f(
π
3

由于:f(x)=
sinx,-
π
2
≤x<0
cosx,0≤x<
π
2

f(
π
3
)=cos
π
3
=
1
2

故答案为:
1
2