题目内容
若α是锐角,且sin(α-| π |
| 6 |
| 1 |
| 3 |
分析:由α是锐角,求出α-
的范围,然后根据sin(α-
)=
的值,利用同角三角函数间的基本关系即可求出cos(α-
)的值,把α变为α-
+
,然后利用两角和的余弦函数公式把所求的式子化简后,把已知sin(α-
)的值和求得的cos(α-
)的值代入即可求出值.
| π |
| 6 |
| π |
| 6 |
| 1 |
| 3 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
解答:解:∵α是锐角,
∴-
<α-
<
,而sin(α-
)=
,∴0<α-
<
,
∴cos(α-
)=
=
,
则cosα=cos[(α-
)+
]
=cos(α-
)cos
-sin(α-
)sin
=
×
-
×
=
.
故答案为:
.
∴-
| π |
| 6 |
| π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| 1 |
| 3 |
| π |
| 6 |
| π |
| 3 |
∴cos(α-
| π |
| 6 |
1-(
|
2
| ||
| 3 |
则cosα=cos[(α-
| π |
| 6 |
| π |
| 6 |
=cos(α-
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=
2
| ||
| 3 |
| ||
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
=
2
| ||
| 6 |
故答案为:
2
| ||
| 6 |
点评:此题考查学生灵活运用同角三角函数间的基本关系、两角和的余弦函数公式及特殊角的三角函数值化简求值,是一道综合题.
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