题目内容
已知函数f(x)=2cos2x+2
sinxcosx.
(1)求函数f(x)定义在[-
,
]上的值域.
(2)在△ABC中,若f(C)=2,2sinB=cos(A-C)-cos(A+C),求tanA的值.
| 3 |
(1)求函数f(x)定义在[-
| π |
| 6 |
| π |
| 3 |
(2)在△ABC中,若f(C)=2,2sinB=cos(A-C)-cos(A+C),求tanA的值.
(1)f(x)=1+cos2x+
sin2x=2sin(2x+
)+1
∵-
≤x≤
∴-
≤2x+
≤
π
∴-
≤sin(2x+
)≤1;
∴f(x)∈[0,3].
即f(x)的值域为[0,3]
(2)由f(C)=2得2sin(2C+
)+1=2,∴sin(2C+
)=
.
∵0<C<π∴
<2C+
<
π
∴2C+
=
∴C=
∴A+B=
.
又∵2sinB=cos(A-C)-cos(A+C)
∴2sinB=2sinAsinC
∴2sin(
-A)=
sinA
即
cosA+sinA=
sinA
∴(
-1)sinA=
cosA
∴tanA=
=
.
| 3 |
| π |
| 6 |
∵-
| π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| 5 |
| 6 |
∴-
| 1 |
| 2 |
| π |
| 6 |
∴f(x)∈[0,3].
即f(x)的值域为[0,3]
(2)由f(C)=2得2sin(2C+
| π |
| 6 |
| π |
| 6 |
| 1 |
| 2 |
∵0<C<π∴
| π |
| 6 |
| π |
| 6 |
| 13 |
| 6 |
∴2C+
| π |
| 6 |
| 5π |
| 6 |
| π |
| 3 |
| 2π |
| 3 |
又∵2sinB=cos(A-C)-cos(A+C)
∴2sinB=2sinAsinC
∴2sin(
| 2π |
| 3 |
| 3 |
即
| 3 |
| 3 |
∴(
| 3 |
| 3 |
∴tanA=
| ||
|
3+
| ||
| 2 |
练习册系列答案
相关题目