题目内容
已知向量
=(
sinx+cosx,1),
=(cosx,-f(x)),
⊥
.
(1)求f(x)的单调区间;
(2)已知A为△ABC的内角,若f(
)=
+
,a=1,b=
,求△ABC的面积.
| m |
| 3 |
| n |
| m |
| n |
(1)求f(x)的单调区间;
(2)已知A为△ABC的内角,若f(
| A |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| 2 |
(1)因为向量
=(
sinx+cosx,1),
=(cosx,-f(x)),
⊥
.
∴f(x)=
sinxcosx+cos2x=
sin2x+
cos2x+
=sin(2x+
)+
,
∴f(x)的单调增区间为:[kπ-
,kπ+
],k∈Z.
函数的单调减区间为[kπ+
,kπ+
],k∈Z.
(2)由f(
)=
+
,a=1,b=
,所以f(
)=sin(A+
)+
=
+
,
∴sin(A+
)=
,
∵A是三角形内角,∴A+
∈(
,
),∴A=
或A=
,
又a=1,b=
,∴A=
,
由正弦定理可得sinB=
=
,?B=
或
,
C=π-A-B=
或
所以△ABC的面积为:
absinC=
sin
=
,
或
absinC=
sin
=
.
| m |
| 3 |
| n |
| m |
| n |
∴f(x)=
| 3 |
| ||
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
∴f(x)的单调增区间为:[kπ-
| π |
| 3 |
| π |
| 6 |
函数的单调减区间为[kπ+
| π |
| 6 |
| 2π |
| 3 |
(2)由f(
| A |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| 2 |
| A |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
∴sin(A+
| π |
| 6 |
| ||
| 2 |
∵A是三角形内角,∴A+
| π |
| 6 |
| π |
| 6 |
| 7π |
| 6 |
| π |
| 6 |
| π |
| 2 |
又a=1,b=
| 2 |
| π |
| 6 |
由正弦定理可得sinB=
| bsinA |
| a |
| ||
| 2 |
| π |
| 4 |
| 3π |
| 4 |
C=π-A-B=
| 7π |
| 12 |
| π |
| 12 |
所以△ABC的面积为:
| 1 |
| 2 |
| ||
| 2 |
| 7π |
| 12 |
1+
| ||
| 4 |
或
| 1 |
| 2 |
| ||
| 2 |
| π |
| 12 |
| ||
| 4 |
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