题目内容
已知sin(x+
)=
,则sin(
-x)+cos2(
-x)=
.
| π |
| 6 |
| 1 |
| 4 |
| 5π |
| 6 |
| π |
| 3 |
| 5 |
| 16 |
| 5 |
| 16 |
分析:利用诱导公式,我们易将sin(
-x)+cos2(
-x)化为sin(x+
)+sin2(x+
),由已知中sin(x+
)=
,代入计算可得结果.
| 5π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 1 |
| 4 |
解答:解:∵sin(x+
)=
,
∴sin(
-x)+cos2(
-x)
=sin[π-(x+
)]+cos2[
-(x+
)]
=sin(x+
)+sin2(x+
)
=
+
=
故答案为:
| π |
| 6 |
| 1 |
| 4 |
∴sin(
| 5π |
| 6 |
| π |
| 3 |
=sin[π-(x+
| π |
| 6 |
| π |
| 2 |
| π |
| 6 |
=sin(x+
| π |
| 6 |
| π |
| 6 |
=
| 1 |
| 4 |
| 1 |
| 16 |
=
| 5 |
| 16 |
故答案为:
| 5 |
| 16 |
点评:本题考查的知识点是三角函数的恒等变换及化简求值,分析已知角与求知角的关系,利用诱导公式,将未知角用已知角表示是解答本题的关键.
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