函数f(x)=cos(-x2)+sin(π-x2)(x∈R).(1)求f(x)的周期;(2)若f(α)=2105,α∈(0,π2),求tan(α+π4)的值.-数学

题目简介

函数f(x)=cos(-x2)+sin(π-x2)(x∈R).(1)求f(x)的周期;(2)若f(α)=2105,α∈(0,π2),求tan(α+π4)的值.-数学

题目详情

函数f(x)=cos(-
x
2
)+sin(π-
x
2
)(x∈R).
(1)求f(x)的周期;
(2)若f(α)=
2
10
5
,α∈(0,
π
2
),求tan(α+
π
4
)的值.
题型:解答题难度:中档来源:不详

答案

(1)由题意知,f(x)=cos(-class="stub"x
2
)+sin(π-class="stub"x
2
)=sinclass="stub"x
2
+cosclass="stub"x
2
=
2
sin(class="stub"x
2
+class="stub"π
4
)

∴f(x)的周期T=class="stub"2π
class="stub"1
2
=4π
(4分)
(2)由f(a)=
2
10
5
代入解析式得,sinclass="stub"α
2
+cosclass="stub"α
2
=
2
10
5

两边平方得:1+sinα=class="stub"8
5
,则sinα=class="stub"3
5

α∈(0,class="stub"π
2
)
,∴cosα=
1-sin2α
=
1-class="stub"9
25
=class="stub"4
5
,(8分)
tanα=class="stub"sinα
cosα
=class="stub"3
4

tan(α+class="stub"π
4
)=
tanα+tanclass="stub"π
4
1-tanαtanclass="stub"π
4
=
class="stub"3
4
+1
1-class="stub"3
4
=7
(12分)

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