问题 解答题
已知f(x)=2sin(
π
4
+
x
2
)sin(
π
4
-
x
2
)+
sin2x
2cosx

(I)若f(α)=
2
2
,α∈(-
π
2
,0),求α的值

(II)若sin
x
2
=
4
5
,x∈(
π
2
,π),求f(x)的值
答案

(I)f(x)=2sin(

π
4
+
x
2
)cos(
π
4
+
x
2
)+
sin2x
2cosx

=sin(

π
2
+x)+sinx=sinx+cosx

=

2
sin(x+
π
4
)

由f(α)=

2
2
,得
2
sin(α+
π
4
)=
2
2

sin(α+

π
4
)=
1
2

α∈(-

π
2
,0)

α+

π
4
∈(-
π
4
π
4
)

α+

π
4
=
π
6
,∴α=-
π
12
(7分)

(II)∵x∈(

π
2
,π),∴
x
2
∈(
π
4
π
2
)

又sin

x
2
=
4
5
,∴cos
x
2
=
3
5

sinx=2sin

x
2
cos
x
2
=
24
25
,cosx=-
1-sin2x
=-
7
25

f(x)=sinx+cosx=

24
25
-
7
25
=
17
25

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