题目内容

已知向量
a
=(cosx,sinx),
b
=(sinx,cosx)
,且x∈[0,
π
2
]

(1)求
a
b
的取值范围;
(2)求证|
a
+
b
|=2sin(x+
π
4
)

(3)求函数f(x)=
a
b
-
2
|
a
+
b
|
的取值范围.
(1)∵
a
b
=sinx•cosx+sinx•cosx=2sinx•cosx=sin2x  (2′)
∵x∈[0,
π
2
],
∴2x∈[0,π]
a
b
∈[0,1](4′)
(2)证明:∵
a
+
b
=(cos+sinx,sinx+cosx)
∴|
a
+
b
|=
2(cosx+sinx)2
(6')
=
2[
2
sin(x+
π
4
)]
2
=2|sin(x+
π
4
)|

∵x∈[0,
π
2
],
∴x+
π
4
∈[
π
4
4
],
∴sin(x+
π
4
)>0,
2|sin(x+
π
4
)|
=2sin(x+
π
4
),
∴|
a
+
b
|=2sin(x+
π
4
).(8')
(3)∵x∈[0,
π
2
],
∴x+
π
4
∈[
π
4
4
]
∴f(x)=
a
b
-
2
|
a
+
b
|

=sin2x-2
2
sin(x+
π
4
)

=2sinxcosx-2(sinx+cosx)(9')
解法1:令t=sinx+cosx
sinx•cosx=
t2-1
2
   (1≤t≤
2

∴y=t2-1-2t(10')
=(t-1)2-2
∴y∈[-2,1-2
2
]
(12')
解法2:f(x)=sin2x-2
2
sin(x+
π
4
)
(9')
=-cos[2(x+
π
4
)]-2
2
sin(x+
π
4
)

=2sin2(x+
π
4
)-2
2
sin(x+
π
4
)
-1(10')
2
2
≤sin(x+
π
4
)
≤1
∴f(x)∈[-2,1-2
2
](12')
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