问题 解答题
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).
答案

(Ⅰ)由sin(x+

π
2
)≠0得x+
π
2
≠kπ,即x≠kπ-
π
2
(k∈Z)

故f(x)的定义域为{x∈R|x≠kπ-

π
2
,k∈Z}.

(Ⅱ)由已知条件得sina=

1-cos2a
=
1-(
3
5
)
2
-
4
5

从而f(a)=

1+
2
cos(2a-
π
4
)
sin(a+
π
2
)

=

1+
2
(cosacos
π
4
+sin2asin
π
4
)
cosa

=

1+cos2a+sina
cosa
=
2cos2a+2sinacosa
cosa

=2(cosa+sina)=

14
5

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