题目内容
已知tan(
+α)=2,tanβ=
.
(1)求tanα的值;
(2)求
的值.
| π |
| 4 |
| 1 |
| 2 |
(1)求tanα的值;
(2)求
| sin(α+β)-2sinαcosβ |
| 2sinαsinβ+cos(α+β) |
(1)∵tan(
+α)=2,
∴tanα=tan[(
+α)-
]=
=
=
.
(2)
=
=
=
=tan(β-α)=
=
=
.
| π |
| 4 |
∴tanα=tan[(
| π |
| 4 |
| π |
| 4 |
tan(
| ||||
1+tan(
|
| 2-1 |
| 1+2×1 |
| 1 |
| 3 |
(2)
| sin(α+β)-2sinαcosβ |
| 2sinαsinβ+cos(α+β) |
| sinαcosβ+cosαsinβ-2sinαcosβ |
| 2sinαsinβ+cosαcosβ-sinαsinβ |
=
| cosαsinβ-sinαcosβ |
| cosαcosβ+sinαsinβ |
| sin(β-α) |
| cos(β-α) |
| tanβ-tanα |
| 1+tanβtanα |
| ||||
1+
|
| 1 |
| 7 |
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