题目内容
已知向量
=(cos
,sin
),
=(2,1),且
⊥
.
(1)求tanθ的值;
(2 )求
的值.
| a |
| θ |
| 2 |
| θ |
| 2 |
| b |
| a |
| b |
(1)求tanθ的值;
(2 )求
| cos2θ | ||||
|
(1)∵
⊥
,向量
=(cos
,sin
),
=(2,1),
即
•
=0
即2cos
+sin
=0,?tan
=-2
∴tanθ=
=
=
. …(5分)
(2)原式=
=
=
=
+1
=
+1=
. …(12分)
| a |
| b |
| a |
| θ |
| 2 |
| θ |
| 2 |
| b |
即
| a |
| b |
即2cos
| θ |
| 2 |
| θ |
| 2 |
| θ |
| 2 |
∴tanθ=
2tan
| ||
1-tan2
|
| 2×(-2) |
| 1-(-2)2 |
| 4 |
| 3 |
(2)原式=
| cos2θ-sin2θ | ||||||||||
|
=
| cosθ+sinθ |
| sinθ |
=
| (cosθ-sinθ)(cosθ+sinθ) |
| (cosθ-sinθ)sinθ |
=
| 1 |
| tanθ |
=
| 3 |
| 4 |
| 7 |
| 4 |
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