题目内容
求函数y=cosx+cos(x-
)(x∈R)的最大值和最小值.
| π | 3 |
分析:将y=cosx+cos(x-
)中的cos(x-
)由两角差的余弦公式展开,再与cosx合并,利用辅助角公式即可求得答案.
| π |
| 3 |
| π |
| 3 |
解答:解:∵y=cosx+cos(x-
)
=cosx+cosxcos
+sinxsin
=
cosx+
sinx
=
(cos
cosx+sin
sinx)
=
cos(x-
),
∵-1≤cos(x-
)≤1,
∴ymax=
,ymin=-
.
| π |
| 3 |
=cosx+cosxcos
| π |
| 3 |
| π |
| 3 |
=
| 3 |
| 2 |
| ||
| 2 |
=
| 3 |
| π |
| 6 |
| π |
| 6 |
=
| 3 |
| π |
| 6 |
∵-1≤cos(x-
| π |
| 6 |
∴ymax=
| 3 |
| 3 |
点评:本题考查三角函数的最值,考查三角函数间关系式,突出辅助角公式的考查,属于中档题.
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