题目内容
已知向量
=(4,5cosα),
=(3,-4tanα)
(1)若
∥
,试求sinα
(2)若
⊥
,且α∈(0,
),求cos(2α-
)的值.
| a |
| b |
(1)若
| a |
| b |
(2)若
| a |
| b |
| π |
| 2 |
| π |
| 4 |
(1)因为向量
=(4,5cosα),
=(3,-4tanα)
由
∥
得,所以15cosα+16tanα=0,即15-15sin2α+16sinα=0,
解得:sinα=
(舍)或sinα=-
.
(2)由
⊥
得,12-20cosα•tanα=0,
∴sinα=
,
又α∈(0,
),∴cosα=
,
sin2α=2sinαcosα=2×
×
=
,cos2α=2cos2α-1=
,
cos(2α-
)=cos2αcos
+sin2αsin
=
.
| a |
| b |
由
| a |
| b |
解得:sinα=
| 5 |
| 3 |
| 3 |
| 5 |
(2)由
| a |
| b |
∴sinα=
| 3 |
| 5 |
又α∈(0,
| π |
| 2 |
| 4 |
| 5 |
sin2α=2sinαcosα=2×
| 3 |
| 5 |
| 4 |
| 5 |
| 24 |
| 25 |
| 7 |
| 25 |
cos(2α-
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 31 |
| 50 |
| 2 |
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