问题 解答题
已知函数f(x)=Asin(2x+φ)(A>0,|φ|<
π
2
),且f(
6
)=-A.
(I)求φ的值;
(Ⅱ)若f(α)=
3
5
A
,f(β+
π
12
)=
5
13
A
π
6
<α<
π
3
,0<β<
π
4
,求cos(2α+2β-
π
6
)的值.
答案

(I)由题意,得f(

6
)=-A⇒Asin(2×
6
+φ)=-A,

∴sin(

3
+φ)=-1,

3
+φ=2kπ+
2
,k∈Z

∵|φ|<

π
2

∴φ=-

π
6

(Ⅱ)由(I)可知,函数f(x)=Asin(2x-

π
6
),

∵f(α)=

3
5
A∴Asin(2α-
π
6
)=
3
5
A

∴sin(2α-

π
6
)=
3
5

π
6
<α<
π
3

π
6
<2α-
π
6
π
3

∴cos(2α-

π
6
)=
4
5

又f(β+

π
12
)=
5
13
A,

∴Asin(2β+

π
6
-
π
6
)=
5
13
A

∴sin2β=

12
13

0<β<

π
4

0<2β<

π
2

∴cos2β=

12
13

∴cos(2α+2β-

π
6
)=cos(2α-
π
6
)cos2β-sin(2α-
π
6
)sin2β

=

4
5
×
12
13
-
3
5
×
5
13

=

33
65

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