3225282
幼苗
共回答了23个问题采纳率:82.6% 举报
f(x)=
a •
b =sin 2 x+sinxcosx=
1-cos2x
2 +
1
2 sin2x=
1
2 (sin2x-cos2x)+
1
2 =
2
2 sin(2x-
π
4 )+
1
2
(1)∵ x∈[0,
π
2 ] ,∴2x-
π
4 ∈[-
π
4 ,
3π
4 ]
∴当2x-
π
4 =-
π
4 ,即x=0时,f(x)最小为-
2
2 ×
2
2 +
1
2 =0
由-
π
2 +2kπ≤2x-
π
4 ≤
π
2 +2kπ,得-
π
8 +kπ≤x≤
3π
8 +kπ,
由
π
2 +2kπ≤2x-
π
4 ≤
3π
2 +2kπ,得
3π
8 +kπ≤x≤
7π
8 +kπ,
取k=0,结合 x∈[0,
π
2 ]
∴函数f(x)的单调增区间为[0,
3π
8 ],单调减区间为[
3π
8 ,
π
2 ]
(2)∵ f(α)=
3
4 ,∴
2
2 sin(2x-
π
4 )+
1
2 =
3
4
∴sin(2x-
π
4 )=
2
4
∵ x∈[0,
π
2 ] ,∴2x-
π
4 ∈[-
π
4 ,
3π
4 ]
∵0<sin(2x-
π
4 )<
1
2
∴2x-
π
4 ∈(0,
π
6 )
∴cos(2x-
π
4 )=
14
4
∴sin2x=sin(2x-
π
4 +
π
4 )=
2
2 sin(2x-
π
4 )+
2
2 cos(2x-
π
4 )=
2
2 (
2
4 +
14
4 )=
7 +1
4
1年前
10