题目内容
函数f(x)=
的值域为( )
| cosx | ||||
cos
|
A、[-
| ||||
B、(-
| ||||
C、[-
| ||||
D、(-
|
分析:函数f(x)=
=cos
+sin
=
sin(
+
),根据-1≤sin(
+
)≤1,以及sin(
+
)≠±1 求得答案.
| cosx | ||||
cos
|
| x |
| 2 |
| x |
| 2 |
| 2 |
| π |
| 4 |
| x |
| 2 |
| π |
| 4 |
| x |
| 2 |
| π |
| 4 |
| x |
| 2 |
解答:解:函数f(x)=
=cos
+sin
=
sin(
+
),
由于-1≤sin(
+
)≤1,故-
≤f(x)≤
.
再根据cos
≠sin
,可得sin(
+
)≠±1,∴-
<f(x)<
.
故选:D.
| cosx | ||||
cos
|
| x |
| 2 |
| x |
| 2 |
| 2 |
| π |
| 4 |
| x |
| 2 |
由于-1≤sin(
| π |
| 4 |
| x |
| 2 |
| 2 |
| 2 |
再根据cos
| x |
| 2 |
| x |
| 2 |
| π |
| 4 |
| x |
| 2 |
| 2 |
| 2 |
故选:D.
点评:本题考查二倍角公式,本题考查两角和的正弦公式的应用,正弦函数的值域,把函数f(x)化为
sin(
+
),是解题的关键,属于中档题.
| 2 |
| π |
| 4 |
| x |
| 2 |
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