问题 解答题
已知向量
a
=(cos
3x
2
,sin
3x
2
)
b
=(cos
x
2
,-sin
x
2
)
x∈[-
π
5
π
2
]

(1)求证:(
a
-
b
)
(
a
+
b
)

(2)|
a
+
b
|=
1
3
,求sin2x的值.
答案

(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
(3分)

(

a
-
b
)•(
a
+
b
)=
a
2
-
b
2
=0(4分)

(

a
-
b
)⊥(
a
+
b
)
(5分)

(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
(8分)

又∵|

a
+
b
|=
1
3

∴cos2x=-

17
18
(10分)

x∈[-

π
5
π
2
]

2x∈[-

5
,π]

∴sin2x=

35
18
(12分)

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