函数f(x)=Asin(ωx-π6)(A>0,ω>0)的最大值为2,其最小正周期为π.(1)求函数f(x)的解析式;(2)设α∈(0,π2),则f(α2)=2,求cosα的值.-数学

题目简介

函数f(x)=Asin(ωx-π6)(A>0,ω>0)的最大值为2,其最小正周期为π.(1)求函数f(x)的解析式;(2)设α∈(0,π2),则f(α2)=2,求cosα的值.-数学

题目详情

函数f(x)=Asin(ωx-
π
6
)
(A>0,ω>0)的最大值为2,其最小正周期为π.
(1)求函数f(x)的解析式;
(2)设α∈(0,
π
2
)
,则f(
α
2
)=
2
,求cosα的值.
题型:解答题难度:中档来源:不详

答案

(1)由题意A=2,…(2分)
∵最小正周期T=π,∴ω=2…(4分)
故函数f(x)的解析式为f(x)=2sin(2x-class="stub"π
6
)
…(5分)
(2)∵f(class="stub"α
2
)=2sin(2×class="stub"α
2
-class="stub"π
6
)=
2
,即sin(α-class="stub"π
6
)=
2
2
,…(6分)
0<α<class="stub"π
2
,∴-class="stub"π
6
<α-class="stub"π
6
<class="stub"π
3
,…(7分)
α-class="stub"π
6
=class="stub"π
4
α=class="stub"π
6
+class="stub"π
4
,…(10分)
cosα=cos(class="stub"π
6
+class="stub"π
4
)=cosclass="stub"π
6
cosclass="stub"π
4
-sinclass="stub"π
6
sinclass="stub"π
4
=
6
-
2
4
…(12分)
(或)∵0<α<class="stub"π
2
,∴-class="stub"π
6
<α-class="stub"π
6
<class="stub"π
3

cos(α-class="stub"π
6
)=
1-sin2(α-class="stub"π
6
)
=
2
2
,…(9分)
cosα=cos[(α-class="stub"π
6
)+class="stub"π
6
]=cos(α-class="stub"π
6
)cosclass="stub"π
6
-sin(α-class="stub"π
6
)sinclass="stub"π
6
=
6
-
2
4
…(12分)

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