题目内容
已知tan(α+
)=
,且-
<α<0,则
=______.
| π |
| 4 |
| 1 |
| 2 |
| π |
| 2 |
| 2sin2α+sin2α | ||
cos(α-
|
∵tan(α+
)=
=
,∴tanα=-
.
再由 tanα=
,sin2α+cos2α=1,-
<α<0,
可得 sinα=-
.
故
=
=
=2
sinα
=2
×(-
)=
.
故答案为
.
| π |
| 4 |
| 1 |
| 2 |
| 1+tanα |
| 1-tanα |
| 1 |
| 3 |
再由 tanα=
| sinα |
| cosα |
| π |
| 2 |
可得 sinα=-
| ||
| 10 |
故
| 2sin2α+sin2α | ||
cos(α-
|
| 2sin α(sinα+cosα) | ||
cos(
|
| 2sin α(sinα+cosα) | ||||
|
| 2 |
=2
| 2 |
| ||
| 10 |
-2
| ||
| 5 |
故答案为
-2
| ||
| 5 |
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