(2007•重庆)已知函数f(x)=1+2cos(2x−π4)sin(x+π2).
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(2007•重庆)已知函数f(x)=
1+
| ||||
sin(x+
|
(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+
2cos(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.
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+
2cos(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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