题目内容
(2010•和平区一模)已知α∈(
,π),且sinα=
;
(Ⅰ)求sin(α+
)的值;
(Ⅱ)求cos(2α+
)的值.
| π |
| 2 |
| ||
| 4 |
(Ⅰ)求sin(α+
| π |
| 4 |
(Ⅱ)求cos(2α+
| π |
| 3 |
分析:(I)先跟据同角三角函数的基本关系以及角的范围求出cosα,然后由两角和与差公式将相应的值代入即可.
(II)由二倍角公式求出sin2α和cos2α,再由两角和与差公式将相应的值代入即可.
(II)由二倍角公式求出sin2α和cos2α,再由两角和与差公式将相应的值代入即可.
解答:解:(I)∵α∈(
,π),且sinα=
∴cosα=-
=-
=-
∴sin(α+
)=sinαcos
+cosαsin
=
×
+(-
)×
=
(II)∵cos2α=cos2α-sin2α=
-
=-
sin2α=2sinαcosα=2×
×(-
)=-
∴cos(2α+
)=cos2αcos
-sin2αsin
=-
×
-(-
)×
=
| π |
| 2 |
| ||
| 4 |
∴cosα=-
| 1-sin2α |
1-
|
| 1 |
| 4 |
∴sin(α+
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
=
| ||
| 4 |
| ||
| 2 |
| 1 |
| 4 |
| ||
| 2 |
=
| ||||
| 8 |
(II)∵cos2α=cos2α-sin2α=
| 1 |
| 16 |
| 15 |
| 16 |
| 7 |
| 8 |
sin2α=2sinαcosα=2×
| ||
| 4 |
| 1 |
| 4 |
| ||
| 8 |
∴cos(2α+
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| 7 |
| 8 |
| 1 |
| 2 |
| ||
| 8 |
| ||
| 2 |
-7+3
| ||
| 16 |
点评:本题主要考查同角三角函数的基本关系的应用,二倍角公式的应用,属于基础题.
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