题目内容
已知cos(α+β)=
,cosβ=
,α,β∈(0,
),求cosα及sin(α+2β)的值.
| 5 |
| 13 |
| 4 |
| 5 |
| π |
| 2 |
∵α,β∈(0,
),∴α+β∈(0,π)
∴sin(α+β)=
=
=
∴sinβ=
=
=
cosα=cos[(α+β)-β]=cos(α+β)cosβ+sin(α+β)sinβ
=
×
+
×
=
sin(α+2β)=sin[(α+β)+β]=sin(α+β)cosβ+cos(α+β)sinβ
=
×
+
×
=
| π |
| 2 |
∴sin(α+β)=
| 1-co s2(α+β) |
1-(
|
| 12 |
| 13 |
∴sinβ=
| 1-cos2β |
1-(
|
| 3 |
| 5 |
cosα=cos[(α+β)-β]=cos(α+β)cosβ+sin(α+β)sinβ
=
| 5 |
| 13 |
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
=
| 56 |
| 65 |
sin(α+2β)=sin[(α+β)+β]=sin(α+β)cosβ+cos(α+β)sinβ
=
| 12 |
| 13 |
| 4 |
| 5 |
| 5 |
| 13 |
| 3 |
| 5 |
=
| 63 |
| 65 |
练习册系列答案
相关题目