题目内容
若cos(x+
)+sinx=
,则cos(2x-
)= .
| π |
| 6 |
| 4 |
| 5 |
| π |
| 3 |
分析:利用两角和与差的余弦可求得cos(x-
)=
,再利用二倍角的余弦即可求得答案.
| π |
| 6 |
| 4 |
| 5 |
解答:解:∵cos(x+
)+sinx
=
cosx-
sinx+sinx
=
cosx+
sinx
=cos(x-
)
=
,
∴cos(2x-
)=2cos2(x-
)-1=2×
-1=
,
故答案为:
.
| π |
| 6 |
=
| ||
| 2 |
| 1 |
| 2 |
=
| ||
| 2 |
| 1 |
| 2 |
=cos(x-
| π |
| 6 |
=
| 4 |
| 5 |
∴cos(2x-
| π |
| 3 |
| π |
| 6 |
| 16 |
| 25 |
| 7 |
| 25 |
故答案为:
| 7 |
| 25 |
点评:本题考查两角和与差的余弦函数,着重考查二倍角的余弦,求得cos(x-
)=
是关键,属于中档题.
| π |
| 6 |
| 4 |
| 5 |
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