在△ABC中,角A、B、C所对的边分别为a、b、c, q =(2a,1), p =(2b

2025-03-01 13:34:12
推荐回答(1个)
回答1:

(I)∵
p
q
,∴2acosC=1×(2b-c),
根据正弦定理,得2sinAcosC=2sinB-sinC,
又∵sinB=sin(A+C)=sinAcosC+cosAsinC,
∴2cosAsinC-sinC=0,即sinC(2cosA-1)=0
∵C是三角形内角,sinC≠0
∴2cosA-1=0,可得cosA=
1
2

∵A是三角形内角,
∴A=
π
3
,得sinA=
3
2
            …(5分)
(II)
-2cos2C
1+tanC
+1
=
2(si n 2 C-co s 2 C)
1+
sinC
cosC
+1
=2cosC(sinC-cosC)+1=sin2C-cos2C,
-2cos2C
1+tanC
+1
=
2
sin(2C-
π
4
),
∵A=
π
3
,得C∈(0,
3
),
∴2C-
π
4
∈(-
π
4
13π
12
),可得-
2
2
<sin(2C-
π
4
)≤1,
∴-1<
2
sin(2C-
π
4
2

即三角函数式
-2cos2C
1+tanC
+1
的取值范围是(-1,
2
].     …(11分)