题目内容
已知A,B,C是三角形△ABC三内角,向量
=(-1,
),
=(cosA,sinA),且
•
=1.
(1)求角A;(2)若tanB=
,求
的值.
| m |
| 3 |
| n |
| m |
| n |
(1)求角A;(2)若tanB=
| 1 |
| 2 |
| 1+sin2B |
| cos2B-sin2B |
(1)∵
•
=1∴(-1,
)•(cosA,sinA)=1,
即
sinA-cosA=1,2(sinA•
-cosA•
)=1,
∴sin(A-
)=
,
∵0<A<π∴-
<A-
<
∴A-
=
∴A=
.
(2)由题知
=
=
=
=3.
| m |
| n |
| 3 |
即
| 3 |
| ||
| 2 |
| 1 |
| 2 |
∴sin(A-
| π |
| 6 |
| 1 |
| 2 |
∵0<A<π∴-
| π |
| 6 |
| π |
| 6 |
| 5π |
| 6 |
∴A-
| π |
| 6 |
| π |
| 6 |
| π |
| 3 |
(2)由题知
| 1+sin2B |
| cos2B-sin2B |
| (sinB+cosB)2 |
| cos2B-sin2B |
| sinB+cosB |
| cosB-sinB |
| 1+tanB |
| 1-tanB |
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