题目内容
已知函数f(x)=2sin(x+
)sin(
-x)+
sin2x
(1)求f(x)的最小正周期.
(2)在锐角△ABC中,角A,B,C对应的边分别为a,b,c,且f(C)=1,c=2,sinB=2sinA,求△ABC的面积S.
| π |
| 4 |
| π |
| 4 |
| 3 |
(1)求f(x)的最小正周期.
(2)在锐角△ABC中,角A,B,C对应的边分别为a,b,c,且f(C)=1,c=2,sinB=2sinA,求△ABC的面积S.
f(x)=2sin(x+
)sin(
-x)+
sin2x
=cos
-cos2x+
sin2x
=2sin(2x-
),
(1)∵ω=2,∴T=
=π;
(2)由f(C)=2sin(2C-
)=1,且C为锐角,
∴C=
,
又sinB=2sinA,根据正弦定理得:b=2a,又c=2,
根据余弦定理c2=a2+b2-2ab•cosC得:a2=
,
则△ABC的面积S=
ab•sinC=
a2=
.
| π |
| 4 |
| π |
| 4 |
| 3 |
=cos
| π |
| 2 |
| 3 |
=2sin(2x-
| π |
| 3 |
(1)∵ω=2,∴T=
| 2π |
| 2 |
(2)由f(C)=2sin(2C-
| π |
| 3 |
∴C=
| π |
| 4 |
又sinB=2sinA,根据正弦定理得:b=2a,又c=2,
根据余弦定理c2=a2+b2-2ab•cosC得:a2=
20-8
| ||
| 17 |
则△ABC的面积S=
| 1 |
| 2 |
| ||
| 2 |
10
| ||
| 17 |
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