题目内容
已知sin(π+α)=-
,计算:
(1)sin(5π-α);
(2)sin(
+α);
(3)cos(α-
);
(4)tan(
-α).
| 1 |
| 2 |
(1)sin(5π-α);
(2)sin(
| π |
| 2 |
(3)cos(α-
| 3π |
| 2 |
(4)tan(
| π |
| 2 |
因为sin(π+α)=-
,所以-sinα=-
,即sinα=
,
(1)sin(5π-α)=sin(π-α)=sinα=
;
(2)sin(
+α)=cosα=±
=±
;
(3)cos(α-
)=cos(
+α)=-sinα=-
;
(4)tan(
-α)=
=
=±
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
(1)sin(5π-α)=sin(π-α)=sinα=
| 1 |
| 2 |
(2)sin(
| π |
| 2 |
| 1-sin2α |
| ||
| 2 |
(3)cos(α-
| 3π |
| 2 |
| π |
| 2 |
| 1 |
| 2 |
(4)tan(
| π |
| 2 |
sin(
| ||
cos(
|
| cosα |
| sinα |
| 3 |
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