题目内容
(1)已知tanα=
,求
的值;
(2)化简:
.
1 |
3 |
1 |
2sinαcosα+cos2α |
(2)化简:
tan(π-α)cos(2π-α)sin(-α+
| ||
cos(-α-π)sin(-π-α) |
(1)∵tanα=
,
∴原式=
=
=
=
.
(2)
=
=
=-1.
1 |
3 |
∴原式=
sin2α+cos2α |
2sinαcosα+cos2α-sin2α |
tan2α+1 |
2tanα+1-tan2α |
| ||||
2×
|
5 |
7 |
(2)
tan(π-α)cos(2π-α)sin(-α+
| ||
cos(-α-π)sin(-π-α) |
-tanαcos(-α)sin(α+
| ||
-cosαsinα |
| ||
-sinα |
练习册系列答案
相关题目