题目内容
锐角α,β满足sinα=
, tanβ=
,则α+β=______.
| ||
| 5 |
| 1 |
| 3 |
∵α为锐角,且sinα=
,
∴cosα=
=
,
又β为锐角,且tanβ=
,
∴cosβ=
=
,
∴sinβ=
=
,
∴cos(α+β)=cosαcosβ-sinαsinβ
=
×
-
×
=
,
又α+β∈(0,π),
则α+β=
.
故答案为:
| ||
| 5 |
∴cosα=
| 1-sin2α |
2
| ||
| 5 |
又β为锐角,且tanβ=
| 1 |
| 3 |
∴cosβ=
|
3
| ||
| 10 |
∴sinβ=
| 1-cos2β |
| ||
| 10 |
∴cos(α+β)=cosαcosβ-sinαsinβ
=
2
| ||
| 5 |
3
| ||
| 10 |
| ||
| 5 |
| ||
| 10 |
| ||
| 2 |
又α+β∈(0,π),
则α+β=
| π |
| 4 |
故答案为:
| π |
| 4 |
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