题目内容
若0<α<
,-
<β<0,cos(
+α)=
,cos(
-
)=
,则cos(α+
)=( )
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| 1 |
| 3 |
| π |
| 4 |
| β |
| 2 |
| ||
| 3 |
| β |
| 2 |
A、
| ||||
B、-
| ||||
C、
| ||||
D、-
|
分析:先利用同角三角函数的基本关系分别求得sin(
+α)和sin(
-
)的值,进而利用cos(α+
)=cos[(
+α)-(
-
)]通过余弦的两角和公式求得答案.
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
| β |
| 2 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
解答:解:∵0<a<
,-
<β<0,
∴
<
+α<
,
<
-
<
∴sin(
+α)=
=
,sin(
-
)=
=
∴cos(α+
)=cos[(
+α)-(
-
)]=cos(
+α)cos(
-
)+sin(
+α)sin(
-
)=
故选C
| π |
| 2 |
| π |
| 2 |
∴
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
| π |
| 2 |
∴sin(
| π |
| 4 |
1-
|
2
| ||
| 3 |
| π |
| 4 |
| β |
| 2 |
1-
|
| ||
| 3 |
∴cos(α+
| β |
| 2 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
5
| ||
| 9 |
故选C
点评:本题主要考查了三角函数的恒等变换及化简求值.关键是根据cos(α+
)=cos[(
+α)-(
-
)],巧妙利用两角和公式进行求解.
| β |
| 2 |
| π |
| 4 |
| π |
| 4 |
| β |
| 2 |
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