问题 解答题
已知向量
a
=(cos
3
2
x
,sin
3
2
x
),
b
=(cos
x
2
,-sin
x
2
),且x∈[0,
π
2
];
(I)求
a
b
及|
a
+
b
|;
(II)若f(x)=
a
b
-
3
|
a
+
b
|sinx,求f(x)的最大值与最小值.
答案

(I)∵向量

a
=(cos
3
2
x
,sin
3
2
x
),
b
=(cos
x
2
,-sin
x
2
),

a
b
=(cos
3
2
x
,sin
3
2
x
)•(cos
x
2
,-sin
x
2
)=cos
3
2
x
•cos
x
2
-sin
3
2
x
sin
x
2
=cos(
3
2
x
+
x
2
)=cos2x,

|

a
|=|
b
|=1

∴|

a
+
b
|2=
a
2
+
b
2
+2
a
b
=2+2cos2x=4cos2x

又∵x∈[0,

π
2
]

∴|

a
+
b
|=2cosx

(II)∵f(x)=

a
b
-
3
|
a
+
b
|sinx=cos2x-2
3
cosxsinx=cos2x-
3
sin2x=2sin(2x+
5
6
π

∵x∈[0,

π
2
],

∴2x+

5
6
π∈[
5
6
π
11
6
π
]

∴当2x+

5
6
π=
5
6
π
,即x=0时,函数取最大值1,

当2x+

5
6
π=
2
,即x=
π
3
时,函数取最小值-2

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