已知函数f(x)=cos2(x?π6)?sin2x.(1)求f(π12)的值;(2)当x∈[0,π2],求函数y=f(x)的值域

2025-06-27 04:14:54
推荐回答(1个)
回答1:

(1)f(x)=

1
2
[1+cos(2x-
π
3
)-(1-cos2x)]=
1
2
[cos(2x-
π
3
)+cos2x]=cos
π
6
cos(2x-
π
6
)=
3
2
cos(2x-
π
6
),
f(
π
12
)
=
3
2
cos(2×
π
12
-
π
6
)=
3
2

(2)∵x∈[0,
π
2
]

∴2x-
π
6
[?
π
6
6
]
,∴-
3
2