问题 解答题
设向量
m
=(cosx,sinx)
,x∈(0,π),
n
=(1 
3
)

(1)若|
m
-
n
|=
5
,求x的值;
(2)设f(x)=(
m
+
n
)•
n
,求函数f(x)的值域.
答案

(1)∵

m
-
n
=(cosx-1 sinx-
3
)

|

m
-
n
|=
5
cos2x-2cosx+1+sin2x-2
3
sinx+3=5

整理得cosx=-

3
sinx

显然cosx≠0∴tanx=-

3
3

∵x∈(0,π),∴x=

6

(2)∵

m
+
n
=(cosx+1 sinx+
3
),

f(x)=(

m
+
n
)•
n
=(cosx+1 sinx+
3
)•(1,
3
)
=cosx+1+
3
sinx+3

=2(

3
2
sinx+
1
2
cosx)+4=2sin(x+
π
6
)+4

∵0<x<π∴

π
6
<x+
π
6
6

-

1
2
<sin(x+
π
6
)≤1⇒-1<2sin(x+
π
6
)≤2

3<2sin(x+

π
6
)+4≤6

即函数f(x)的值域为(3,6].

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