题目内容
已知cos(
+x)=
,且0<x<
,求
+sin(2x-
)的值.
| π |
| 4 |
| 3 |
| 5 |
| π |
| 4 |
sin(
| ||
| cos(2x+5π) |
| 3π |
| 2 |
分析:先根据同角三角函数之间的关系求出sin(x+
);再借助于诱导公式对所求问题进行化简整理到用sin(x+
)以及cos(
+x)=
,表示的形式即可得到答案.
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 3 |
| 5 |
解答:解:因为cos(
+x)=
,且0<x<
,
所以:sin(x+
)=
.
所以:
+sin(2x-
)
=
+sin(2x+
)
=-
+sin(2x+
)
=-
+2sin(x+
)cos(x+
)
=-
+2×
×
=-
+
=
.
| π |
| 4 |
| 3 |
| 5 |
| π |
| 4 |
所以:sin(x+
| π |
| 4 |
| 4 |
| 5 |
所以:
sin(
| ||
| cos(2x+5π) |
| 3π |
| 2 |
=
cos(x+
| ||
| -cos2x |
| π |
| 2 |
=-
cos(x+
| ||
sin(2x+
|
| π |
| 2 |
=-
cos(x+
| ||||
2sin(x+
|
| π |
| 4 |
| π |
| 4 |
=-
| ||||
2×
|
| 3 |
| 5 |
| 4 |
| 5 |
=-
| 5 |
| 8 |
| 24 |
| 25 |
=
| 67 |
| 200 |
点评:本题主要考查三角函数的恒等变换及化简求值.解决问题的关键在于对公式的熟练掌握以及灵活运用.
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