题目内容
已知函数f(x)=tan(3x+
)
(Ⅰ)求f(
)的值;
(Ⅱ)若α∈(π,2π),且f(
)=2,求cos(α-
)的值.
| π |
| 4 |
(Ⅰ)求f(
| π |
| 9 |
(Ⅱ)若α∈(π,2π),且f(
| α |
| 3 |
| π |
| 4 |
(Ⅰ)f(
)=tan(
+
)=
=
=-2-
(6分)
(Ⅱ)由f(
)=2得tanα=
,(8分)
由题可知α是第三象限角.sinα=-
,cosα=-
(10分)
故cos(α-
)=-
(12分).
| π |
| 9 |
| π |
| 3 |
| π |
| 4 |
tan
| ||||
1-tan
|
| ||
1-
|
| 3 |
(Ⅱ)由f(
| α |
| 3 |
| 1 |
| 3 |
由题可知α是第三象限角.sinα=-
| 1 | ||
|
| 3 | ||
|
故cos(α-
| π |
| 4 |
2
| ||
| 5 |
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