设φ∈(0,
(Ⅰ)求φ的值; (Ⅱ)若x∈[0,
|
(Ⅰ)∵f(
)=sin2(π 4
+φ)=π 4
[1-cos(1 2
+2φ)]=π 2
(1+sin2φ)=1 2
,∴sin2φ=3 4
(4分)1 2
∵φ∈(0,
),∴2φ∈(0,π 4
),∴2φ=π 2
,φ=π 6
.(6分)π 12
(Ⅱ)由(Ⅰ)得f(x)=sin2(x+
)=-π 12
cos(2x+1 2
)+π 6
(8分)1 2
∵0≤x≤
,∴π 2
≤2x+π 6
≤π 6
(9分)7π 6
当2x+
=π,即x=π 6
时,cos(2x+5π 12
)取得最小值-1(11分)π 6
∴f(x)在[0,
]上的最大值为1,此时x=π 2
(12分)5π 12