已知函数f(x)=2sin(13x+φ)(x∈R,-π2<φ<0)图象上一个最低点M(-π,-2)(Ⅰ)求f(x)的解析式;(Ⅱ)设α,β∈[0,π2],f(3α+π2)=1013,f(3β+2π)=

题目简介

已知函数f(x)=2sin(13x+φ)(x∈R,-π2<φ<0)图象上一个最低点M(-π,-2)(Ⅰ)求f(x)的解析式;(Ⅱ)设α,β∈[0,π2],f(3α+π2)=1013,f(3β+2π)=

题目详情

已知函数f(x)=2sin(
1
3
x+φ)(x∈R,-
π
2
<φ<0)图象上一个最低点M(-π,-2)
(Ⅰ)求f(x)的解析式;
(Ⅱ)设α,β∈[0,
π
2
],f(3α+
π
2
)=
10
13
,f(3β+2π)=
6
5
,求cos(α+β)的值.
题型:解答题难度:中档来源:不详

答案

(I)把点M(-π,-2)代入得-2=2sin(class="stub"1
3
×(-π)+φ)

sin(φ-class="stub"π
3
)=-1
,∵-class="stub"π
2
<φ<0
,∴-class="stub"5π
6
<φ-class="stub"π
3
<-class="stub"π
3

φ-class="stub"π
3
=-class="stub"π
2
,解得φ=-class="stub"π
6

f(x)=2sin(class="stub"1
3
x-class="stub"π
6
)

(II)f(3α+class="stub"π
2
)=2sin[class="stub"1
3
(3α+class="stub"π
2
)-class="stub"π
6
]
=2sinα=class="stub"10
13
,∴sinα=class="stub"5
13

α∈[0,class="stub"π
2
]
,∴cosα=
1-sin2α
=class="stub"12
13

f(3β+2π)=2sin[class="stub"1
3
(3β+2π)-class="stub"π
6
]
=2sin(β+class="stub"π
2
)
=2cosβ=class="stub"6
5

cosβ=class="stub"3
5
,∵β∈[0,class="stub"π
2
]
,∴sinβ=class="stub"4
5

∴cos(α+β)=cosαcosβ-sinαsinβ=class="stub"12
13
×class="stub"3
5
-class="stub"5
13
×class="stub"4
5
=class="stub"16
65

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