题目内容
已知向量
=(1,cos
)与
=(
sin
+cos
,y)共线,且有函数y=f(x).
(Ⅰ)若f(x)=1,求cos(
-2x)的值;
(Ⅱ)在△ABC中,角A,B,C,的对边分别是a,b,c,且满足2acosC+c=2b,求函数f(B)的取值范围.
| a |
| x |
| 2 |
| b |
| 3 |
| x |
| 2 |
| x |
| 2 |
(Ⅰ)若f(x)=1,求cos(
| 2π |
| 3 |
(Ⅱ)在△ABC中,角A,B,C,的对边分别是a,b,c,且满足2acosC+c=2b,求函数f(B)的取值范围.
(Ⅰ)∵
与
共线,∴
=
,y=
sin
cos
+cos2
=
sinx+
(1+cosx)=sin(x+
)+
,∴f(x)=sin(x+
)+
=1,
即sin(x+
)=
,∴cos(
-2x)=cos2(
-x)=2cos2(
-x)-1=2sin2(x+
)-1=-
.
(Ⅱ)已知2acosC+c=2b,
由正弦定理得:
,
,∴cosA=
,∴在△ABC中∠A=
,f(B)=sin(B+
)+
.∵∠A=
,∴0<B<
,
<B+
<
,
∴
<sin(B+
)≤1,1<f(B)≤
,∴函数f(B)的取值范围为(1,
].
| a |
| b |
| 1 | ||||||
|
cos
| ||
| y |
| 3 |
| x |
| 2 |
| x |
| 2 |
| x |
| 2 |
| ||
| 2 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
即sin(x+
| π |
| 6 |
| 1 |
| 2 |
| 2π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 6 |
| 1 |
| 2 |
(Ⅱ)已知2acosC+c=2b,
由正弦定理得:
|
|
| 1 |
| 2 |
| π |
| 3 |
| π |
| 6 |
| 1 |
| 2 |
| π |
| 3 |
| 2π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| 5π |
| 6 |
∴
| 1 |
| 2 |
| π |
| 6 |
| 3 |
| 2 |
| 3 |
| 2 |
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