问题 解答题
在△ABC中,∠A、∠B、∠C所对的边分别为a、b、c,且满足cos
A
2
=
2
5
5
AB
AC
=3,b+c=6
(I)求a的值;
(II)求
2sin(A+
π
4
)sin(B+C+
π
4
)
1-cos2A
的值.
答案

(I)∵cos

A
2
=
2
5
5

∴cosA=2cos2

A
2
-1=
3
5

AB
AC
=3,即bccosA=3,

∴bc=5,又b+C=6,

∴b=5,c=1或b=1,c=5,

由余弦定理得:a2=b2+c2-2bccosA=20,

∴a=2

5

(II)

2sin(A+
π
4
)sin(B+C+
π
4
)
1-cos2A

=

2sin(A+
π
4
)sin(π-A+
π
4
)
1-cos2A
=
2sin(A+
π
4
)sin(A-
π
4
)
1-cos2A

=

2sin(A+
π
4
)cos[
π
2
-(A-
π
4
)]
1-cos2A
=
-2sin(A+
π
4
)cos(A+
π
4
)
1-cos2A

=-

sin(2A+
π
2
)
1-cos2A
=-
cos2A
1-cos2A

∴cosA=

3
5
,∴cos2A=2cos2A-1=-
7
25

∴原式=-

-
7
25
1+
7
25
=
7
32

选择题
多项选择题