题目内容

已知向量
a
=(cos
3x
2
,sin
3x
2
)
b
=(cos
x
2
,-sin
x
2
)
x∈[-
π
3
π
2
]

(1)求证:(
a
-
b
)
(
a
+
b
)

(2)|
a
+
b
|=
1
3
,求cosx的值.
(1)∵
a
=(cos
3x
2
,sin
3x
2
)
b
=(cos
x
2
,-sin
x
2
)

a
2
=cos2
3x
2
+sin2
3x
2
=1
b
2
=cos2
x
2
+sin2
x
2
=1

(
a
-
b
)•(
a
+
b
)=
a
2
-
b
2
=0

(
a
-
b
)
(
a
+
b
)


(2)∵|
a
+
b
|=
(
a
+
b
)
2
=
a
2
+2
a
b
+
b
2

=
1+2(cos
3x
2
•cos
x
2
+sin
3x
2
•sin
x
2
)+1

=
2+2cos2x

=
1
3

∴2+2cos2x=
1
9
,即cos2x=-
17
18

2cos2x-1=-
17
18

cos2x=
1
36

x∈[-
π
3
π
2
]

cosx=
1
6
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