题目内容
已知sin(
-α)=
(0<α<
),则cosα=
.
| π |
| 4 |
| ||
| 10 |
| π |
| 2 |
| 4 |
| 5 |
| 4 |
| 5 |
分析:先计算出cos(
-α)=
,再根据cosα=cos[
-(
-α)]=cos
cos(
-α)+sin
sin(
-α),即可求得结论.
| π |
| 4 |
7
| ||
| 10 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
解答:解:∵0<α<
,∴-
<
-α<
∵sin(
-α)=
,∴cos(
-α)=
∴cosα=cos[
-(
-α)]=cos
cos(
-α)+sin
sin(
-α)=
×
+
×
=
故答案为:
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
∵sin(
| π |
| 4 |
| ||
| 10 |
| π |
| 4 |
7
| ||
| 10 |
∴cosα=cos[
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| ||
| 2 |
7
| ||
| 10 |
| ||
| 2 |
| ||
| 10 |
| 4 |
| 5 |
故答案为:
| 4 |
| 5 |
点评:本题考查三角恒等变换,考查学生分析解决问题的能力,属于中档题.
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