题目内容
设f(x)=sinx+cosx,若| π |
| 4 |
| π |
| 2 |
分析:利用两角和的正弦公式可得f(x)=
sin(x+
),由
<x1<x2<
,得
<x1+
<x2+
<
,
故sin(x1+
)>sin(x2+
),从而得到f(x1)>f(x2).
| 2 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
故sin(x1+
| π |
| 4 |
| π |
| 4 |
解答:解:∵f(x)=sinx+cosx=
sin(x+
),若
<x1<x2<
,则
<x1+
<x2+
<
,
∴sin(x1+
)>sin(x2+
),∴f(x1)>f(x2),
故答案为:f(x1)>f(x2).
| 2 |
| π |
| 4 |
| π |
| 4 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
∴sin(x1+
| π |
| 4 |
| π |
| 4 |
故答案为:f(x1)>f(x2).
点评:本题考查两角和的正弦公式,正弦函数的单调性,判断
<x1+
<x2+
<
,是解题的关键.
| π |
| 2 |
| π |
| 4 |
| π |
| 4 |
| 3π |
| 4 |
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