题目内容
已知cos(θ-
)=
,
<θ<
,求cosθ.
| π |
| 6 |
| 12 |
| 13 |
| π |
| 6 |
| π |
| 2 |
∵
<θ<
,
∴0<θ-
<
,又cos(θ-
)=
,
∴sin(θ-
)=
=
,
则cosθ=cos[(θ-
)+
]
=cos(θ-
)cos
-sin(θ-
)sin
=
×
-
×
=
.
| π |
| 6 |
| π |
| 2 |
∴0<θ-
| π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| 12 |
| 13 |
∴sin(θ-
| π |
| 6 |
1-cos2(θ-
|
| 5 |
| 13 |
则cosθ=cos[(θ-
| π |
| 6 |
| π |
| 6 |
=cos(θ-
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=
| 12 |
| 13 |
| ||
| 2 |
| 5 |
| 13 |
| 1 |
| 2 |
=
12
| ||
| 26 |
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