已知函数f(x)=cos(-x2)+cos(4k+12π-x2),k∈Z,x∈R.(1)求f(x)的最小正周期;(2)求f(x)在[0,π)上的减区间;(3)若f(α)=2105,α∈(0,π2),求

题目简介

已知函数f(x)=cos(-x2)+cos(4k+12π-x2),k∈Z,x∈R.(1)求f(x)的最小正周期;(2)求f(x)在[0,π)上的减区间;(3)若f(α)=2105,α∈(0,π2),求

题目详情

已知函数f(x)=cos(-
x
2
)+cos(
4k+1
2
π-
x
2
),k∈Z,x∈R.
(1)求f(x)的最小正周期;
(2)求f(x)在[0,π)上的减区间;
(3)若f(α)=
2
10
5
,α∈(0,
π
2
),求tan(2α+
π
4
)的值.
题型:解答题难度:中档来源:无为县模拟

答案

(1)f(x)=cos(-class="stub"1
2
π
)+cos(class="stub"4k+1
2
π-class="stub"1
2
x

=cosclass="stub"1
2
x
+cos(2kπ+class="stub"1
2
π-class="stub"1
2
x

=sinclass="stub"1
2
x
+cosclass="stub"1
2
x
=
2
sin(class="stub"1
2
x
+class="stub"π
4
),
所以,f(x)的最小正周期T=class="stub"2π
class="stub"1
2
=4π                
(2)由class="stub"1
2
π
+2kπ≤class="stub"1
2
x+class="stub"π
4
≤class="stub"3π
2
+2kπ
,k∈Z
class="stub"1
2
π+4kπ≤x≤class="stub"5π
2
+4kπ
,k∈z
令k=0,得class="stub"π
2
≤x≤class="stub"5π
2

令k=-1可得,-class="stub"7π
12
≤x≤-class="stub"3π
2

∵x∈(0,class="stub"1
2
π)

∴f(x)在(0,π)上的单调递减区间是[class="stub"1
2
π,π

(3)由f(α)=
2
10
5
可得sinclass="stub"α
2
+cosclass="stub"α
2
=
2
10
5

两边同时平方可得,1+sinα=class="stub"8
5

∴sinα=class="stub"3
5

α∈(0,class="stub"1
2
π)

∴cosα=class="stub"4
5

tanα=class="stub"sinα
cosα
=class="stub"3
4
,tan2α=class="stub"2tanα
1-tan2α
=
2×class="stub"3
4
1-class="stub"9
16
=class="stub"24
7

∴tan(2α+class="stub"π
4
)=class="stub"1+tan2α
1-tan2α
=
1+class="stub"24
7
1-class="stub"24
7
=-class="stub"31
17

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