题目内容
已知cos(α+
)=
，
≤α＜
，求cos(2α+
)的值．
| π |
| 4 |
| 3 |
| 5 |
| π |
| 2 |
| 3π |
| 2 |
| π |
| 4 |
cos(2α+
)=cos2αcos
-sin2αsin
=
(cos2α-sin2α).
≤α+
且cos(α+
)＞0，
≤α+
＜
，
∴sin(α+
)=-
=-
从而cos2α=sin(2α+
)=2sin(α+
)cos(α+
)=-
，
sin2α=-cos(2α+
)=1-2cos2(α+
)=
∴cos(2α+
)=
×(-
-
)=-
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| ||
| 2 |
| 3π |
| 4 |
| 7π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| 7π |
| 4 |
∴sin(α+
| π |
| 4 |
1-cos2(α+
|
| 4 |
| 5 |
从而cos2α=sin(2α+
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 24 |
| 25 |
sin2α=-cos(2α+
| π |
| 2 |
| π |
| 4 |
| 7 |
| 25 |
∴cos(2α+
| π |
| 4 |
| ||
| 2 |
| 24 |
| 25 |
| 7 |
| 25 |
31
| ||
| 50 |
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| π |
| 4 |
| π |
| 4 |
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