题目内容
已知cos(15°+α)=
,α为锐角,求:
.
| 3 |
| 5 |
| tαn(435°-α)+sin(α-165°) |
| cos(195°+α)×sin(105°+α) |
考点:运用诱导公式化简求值
专题:三角函数的求值
分析:根据cos(15°+α)=
,α为锐角,求得sin(15°+α)和cot(15°+α)的值,再利用诱导公式把要求的式子化为
,计算求得结果.
| 3 |
| 5 |
| cot(15°+α)-sin(15°+α) |
| -cos2(15°+α) |
解答:
解:∵cos(15°+α)=
,α为锐角,
∴sin(15°+α)=
,
∴cot(15°+α)=
=
.
∴
=
=
=
=
-
=
.
| 3 |
| 5 |
∴sin(15°+α)=
| 4 |
| 5 |
∴cot(15°+α)=
| cos(15°+α) |
| sin(15°+α) |
| 3 |
| 4 |
∴
| tαn(435°-α)+sin(α-165°) |
| cos(195°+α)×sin(105°+α) |
| tan(75°-α)-sin(α+15°) |
| -cos(15°+α)•cos(15°+α) |
=
| cot(15°+α)-sin(15°+α) |
| -cos2(15°+α) |
| ||||
-(
|
| 20 |
| 9 |
| 25 |
| 12 |
| 5 |
| 36 |
点评:本题主要考查同角三角函数的基本关系、诱导公式的应用,属于中档题.
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