题目内容
已知0<α<π,tanα=-2.
(1)求sin(α+
)的值;
(2)求
的值;
(3)2sin2α-sinαcosα+cos2α
(1)求sin(α+
| π |
| 6 |
(2)求
2cos(
| ||
sin(
|
(3)2sin2α-sinαcosα+cos2α
因为0<α<π,tanα=-2,所以sinα=
,cosα=
(1)sin(α+
)=sinαcos
+cosαsin
=
+(
)×
=
(2)原式=
=
=-1
(3)原式=
=
=
2
| ||
| 5 |
-
| ||
| 5 |
(1)sin(α+
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
2
| ||
| 5 |
| ||
| 2 |
-
| ||
| 5 |
| 1 |
| 2 |
2
| ||||
| 10 |
(2)原式=
| -2sinα+cosα |
| cosα+3sinα |
| -2tanα+1 |
| 1+3tanα |
(3)原式=
| 2sin2α-sinαcosα+cos2α |
| sin2α+cos2a |
=
| 2tan2α-tanα +1 |
| tan2α+1 |
| 11 |
| 5 |
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