已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R)(1)求函数f(x)的最小正周期;(2)求使函数f(x)取得最大值的x集合;(3)若θ∈(0,π2),且f(θ)=53,求c

题目简介

已知函数f(x)=3sin(2x-π6)+2sin2(x-π12)(x∈R)(1)求函数f(x)的最小正周期;(2)求使函数f(x)取得最大值的x集合;(3)若θ∈(0,π2),且f(θ)=53,求c

题目详情

已知函数f(x)=
3
sin(2x-
π
6
)+2sin2(x-
π
12
)(x∈R)

(1)求函数f(x)的最小正周期;
(2)求使函数f(x)取得最大值的x集合;
(3)若θ∈(0,
π
2
)
,且f(θ)=
5
3
,求cos4θ的值.
题型:解答题难度:中档来源:不详

答案

(1)∵sin2(x-class="stub"π
12
)=class="stub"1
2
[1-cos2(x-class="stub"π
12
)]=class="stub"1
2
-class="stub"1
2
cos(2x-class="stub"π
6

∴f(x)=
3
sin(2x-class="stub"π
6
)+[1-cos(2x-class="stub"π
6
)]
=2[sin(2x-class="stub"π
6
)cosclass="stub"π
6
-cos(2x-class="stub"π
6
)sinclass="stub"π
6
]+1
=2sin(2x-class="stub"π
3
)+1
由此可得函数f(x)的最小正周期T=class="stub"2π
2

(2)∵x∈R,∴当2x-class="stub"π
3
=class="stub"π
2
+2kπ(k∈Z)时,函数有最大值为3
解之得x=class="stub"5π
12
+kπ(k∈Z),
得f(x)取得最大值的x集合为{x|x=class="stub"5π
12
+kπ(k∈Z)}
(3)f(θ)=class="stub"5
3
即2sin(2θ-class="stub"π
3
)+1=class="stub"5
3

解之得sin(2θ-class="stub"π
3
)=class="stub"1
3

θ∈(0,class="stub"π
2
)
,得2θ-class="stub"π
3
∈(-class="stub"π
3
class="stub"2π
3

∴根据sin(2θ-class="stub"π
3
)=class="stub"1
3
<class="stub"1
2
,得2θ-class="stub"π
3
∈(0,class="stub"π
6

因此cos(2θ-class="stub"π
3
)=
1-(class="stub"1
3
)2
=
2
2
3

∴cos2θ=cos[(2θ-class="stub"π
3
)+class="stub"π
3
]=
2
2
3
×class="stub"1
2
-class="stub"1
3
×
3
2
=
2
2
-
3
6

cos4θ=2cos22θ-1=2(
2
2
-
3
6
)2-1=
-7-4
6
18

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