已知函数f(x)=2sin(ωx+φ)(ω>0,-π2<φ<π2)的图象如图所示,直线x=3π8,x=7π8是其两条对称轴.(1)求函数f(x)的解析式;(2)若f(a)=65,且π8<α<3π8,求

题目简介

已知函数f(x)=2sin(ωx+φ)(ω>0,-π2<φ<π2)的图象如图所示,直线x=3π8,x=7π8是其两条对称轴.(1)求函数f(x)的解析式;(2)若f(a)=65,且π8<α<3π8,求

题目详情

已知函数f(x)=2sin(ωx+φ)(ω>0,-
π
2
φ<
π
2
)的图象如图所示,直线x=
8
,x=
8
是其两条对称轴.
(1)求函数f(x)的解析式;
(2)若f(a)=
6
5
,且
π
8
<α<
8
,求f(
π
8
)的值.360优课网
题型:解答题难度:中档来源:不详

答案

(本题满分14分)
(1)由题意,class="stub"T
2
=class="stub"7π
8
-class="stub"3π
8
=class="stub"π
2
,∴T=π.
又ω>0,故ω=2,∴f(x)=2sin(2x+φ).(2分)
由f(class="stub"3π
8
)=2sin(class="stub"3π
4
+φ)=2,解得φ=2kπ-class="stub"π
4
(k∈Z).
又-class="stub"π
2
<φ<class="stub"π
2
,∴φ=-class="stub"π
4
,∴f(x)=2sin(2x-class="stub"π
4
).(5分)
由2kπ-class="stub"π
2
≤2x-class="stub"π
4
≤2kπ+class="stub"π
2
(k∈Z),知kπ-class="stub"π
8
≤x≤kπ+class="stub"3π
8
(k∈Z),
∴函数f(x)的单调增区间为[kπ-class="stub"π
8
,kπ+class="stub"3π
8
](k∈Z).(7分)
(2)解法1:依题意得2sin(2α-class="stub"π
4
)=class="stub"6
5
,即sin(2α-class="stub"π
4
)=class="stub"3
5
,(8分)
class="stub"π
8
<α<class="stub"3π
8
,∴0<2α-class="stub"π
4
class="stub"π
2

∴cos(2α-class="stub"π
4
)=class="stub"4
5
,(10分)
f(class="stub"π
8
+α)=2sin[(2α-class="stub"π
4
)+class="stub"π
4
].
∵sin[(2α-class="stub"π
4
)+class="stub"π
4
]=sin(2α-class="stub"π
4
)cosclass="stub"π
4
+cos(2α-class="stub"π
4
)sinclass="stub"π
4
=
2
2
class="stub"3
5
+class="stub"4
5
)=
7
2
10

∴f(class="stub"π
8
+α)=
7
2
5
.(14分)
解法2:依题意得sin(2α-class="stub"π
4
)=class="stub"3
5
,得sin2α-cos2α=
3
2
5
,①(9分)
class="stub"π
8
<α<class="stub"3π
8
,∴0<2α-class="stub"π
4
class="stub"π
2

∴cos(α-class="stub"π
4
)=class="stub"4
5
,(11分)
由cos(2α-class="stub"π
4
)=class="stub"4
5
得sin2α+cos2α=
4
2
5
.②
①+②得2sin2α=
7
2
5

∴f(class="stub"π
8
+α)=
7
2
5
.(14分)

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