题目内容
在△ABC中,三个内角分别为A,B,C,且sin(B+
)=2cosB.
(1)若cosC=
,AC=3,求A、B.
(2)若A∈(0,
),且cos(B-A)=
,求sinA.
| π |
| 6 |
(1)若cosC=
| ||
| 3 |
(2)若A∈(0,
| π |
| 3 |
| 4 |
| 5 |
(1)在△ABC中,由sin(B+
)=2cosB 可得 sinB×
+cosB×
=2cosB,∴sinB=
cosB,∴tanB=
,B=
.
由cosC=
,得 sinC=
.
∴cosA=-cos(B+C)=-cosBcosC+sinBsinC=-
×
+
×
=
,∴A=arccos
.
(2)若A∈(0,
),且cos(B-A)=
,则有 cos(
-A)=
,∴sin(
-A)=
.
∴sin(-A)=sin[(
-A)-
)=sin(
-A)cos
-cos(
-A)sin
=
×
-
×
=
,
故 sinA=
.
| π |
| 6 |
| ||
| 2 |
| 1 |
| 2 |
| 3 |
| 3 |
| π |
| 3 |
由cosC=
| ||
| 3 |
| ||
| 3 |
∴cosA=-cos(B+C)=-cosBcosC+sinBsinC=-
| 1 |
| 2 |
| ||
| 3 |
| ||
| 2 |
| ||
| 3 |
3-
| ||
| 3 |
3-
| ||
| 3 |
(2)若A∈(0,
| π |
| 3 |
| 4 |
| 5 |
| π |
| 3 |
| 4 |
| 5 |
| π |
| 3 |
| 3 |
| 5 |
∴sin(-A)=sin[(
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 3 |
| 5 |
| 1 |
| 2 |
| 4 |
| 5 |
| ||
| 2 |
3-4
| ||
| 10 |
故 sinA=
-3+4
| ||
| 10 |
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