已知函数f(x)=1+2cos(2x-π4)sin(x+π2).(Ⅰ)求f(x)的定义域;(Ⅱ)若角α在第一象限且cosα=35,求f(α).-数学

题目简介

已知函数f(x)=1+2cos(2x-π4)sin(x+π2).(Ⅰ)求f(x)的定义域;(Ⅱ)若角α在第一象限且cosα=35,求f(α).-数学

题目详情

已知函数f(x)=
1+
2
cos(2x-
π
4
)
sin(x+
π
2
)

(Ⅰ)求f(x)的定义域;
(Ⅱ)若角α在第一象限且cosα=
3
5
,求f(α).
题型:解答题难度:中档来源:重庆

答案

(Ⅰ)由sin(x+class="stub"π
2
)
≠0得x+class="stub"π
2
≠kπ,即x≠kπ-class="stub"π
2
(k∈Z)

故f(x)的定义域为{x∈R|x≠kπ-class="stub"π
2
,k∈Z}

(Ⅱ)由已知条件得sina=
1-cos2a
=
1-(class="stub"3
5
)
2
-class="stub"4
5

从而f(a)=
1+
2
cos(2a-class="stub"π
4
)
sin(a+class="stub"π
2
)

=
1+
2
(cosacosclass="stub"π
4
+sin2asinclass="stub"π
4
)
cosa

=class="stub"1+cos2a+sina
cosa
=
2cos2a+2sinacosa
cosa

=2(cosa+sina)=class="stub"14
5

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