题目内容
若cos(α-
)=
,α∈(
,2π),则cosα的值为( )
| π |
| 4 |
-
| ||
| 5 |
| 3π |
| 2 |
A.
| B.-
| C.-
| D.
|
cos(α-
)=
,α∈(
,2π),α-
∈(
,
)
sin(α-
)=-
=-
cosα=cos[(α-
)+
]=cos(α-
)cos
-sin(α-
)sin
=-
×
-(-
)×
=
故选D
| π |
| 4 |
-
| ||
| 5 |
| 3π |
| 2 |
| π |
| 4 |
| 5π |
| 4 |
| 7π |
| 4 |
sin(α-
| π |
| 4 |
1-(-
|
2
| ||
| 5 |
cosα=cos[(α-
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
=-
| ||
| 5 |
| ||
| 2 |
2
| ||
| 5 |
| ||
| 2 |
=
| ||
| 10 |
故选D
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