已知函数f(x)=Asin(ωx+ϕ),(A>0,ω>0,0<ϕ<π2)图象关于点B(-π4,0)对称,点B到函数y=f(x)图象的对称轴的最短距离为π2,且f(π2)=1.(1)求A,ω,ϕ的值;(

题目简介

已知函数f(x)=Asin(ωx+ϕ),(A>0,ω>0,0<ϕ<π2)图象关于点B(-π4,0)对称,点B到函数y=f(x)图象的对称轴的最短距离为π2,且f(π2)=1.(1)求A,ω,ϕ的值;(

题目详情

已知函数f(x)=Asin(ωx+ϕ),(A>0,ω>0,0<ϕ<
π
2
)
图象关于点B(-
π
4
,0)
对称,点B到函数y=f(x)图象的对称轴的最短距离为
π
2
,且f(
π
2
)=1

(1)求A,ω,ϕ的值;
(2)若0<θ<π,且f(θ)=
1
3
,求cos2θ
的值.
题型:解答题难度:中档来源:宁波模拟

答案

(1)∵点B到函数y=f(x)图象的对称轴的最短距离为class="stub"π
2
,且点B是函数f(x)=Asin(ωx+ϕ),(A>0,ω>0,0<ϕ<class="stub"π
2
)
的对称中心
class="stub"T
4
=class="stub"π
2
,∴T=2π
class="stub"2π
ω
=4×class="stub"π
2
=2π,
∴ω=1
又∵点B(-class="stub"π
4
,0)
是函数f(x)的对称中心
f(-class="stub"π
4
)=Asin(-class="stub"π
4
+ϕ)=0

sin(ϕ-class="stub"π
4
)=0

∵0<ϕ<class="stub"π
2

∴-class="stub"π
4
<ϕ-class="stub"π
4
<class="stub"π
4

∴ϕ-class="stub"π
4
=0,
∴ϕ=class="stub"π
4

f(class="stub"π
2
)=Asin(class="stub"π
2
+class="stub"π
4
)=
2
2
A=1,
∴A=
2

∴A=
2
,ω=1,ϕ=class="stub"π
4

(2)∵f(θ)=
2
sin(θ+class="stub"π
4
)=sinθ+cosθ
=class="stub"1
3

∴(sinθ+cosθ)2=1+2sinθcosθ=class="stub"1
9

∴2sinθcosθ=-class="stub"8
9
<0,∵0<θ<π
∴sinθ>0,
∴cosθ<0
∴sinθ-cosθ=
(sinθ-cosθ) 2
=
1-2sinθcosθ
=
1+class="stub"8
9
=
17
3

∴cos2θ=(cosθ+sinθ)(cosθ-sinθ)=class="stub"1
3
×(-
17
3
)=-
17
9

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