问题 解答题
已知函数f(x)=cos2x+sin2x
(1)求f(x)的最大值和最小正周期;
(2)设α,β∈[0,
π
2
]
,f(
α
2
+
π
8
)=
5
2
,f(
β
2
)=
2
,求sin(α+β)的值.
答案

(1)∵f(x)=cos2x+sin2x=

2
2
2
cos2x+
2
2
sin2x)=
2
sin(2x+
π
4
),

∵-1≤sin(2x+

π
4
)≤1,

∴f(x)的最大值为

2

∵ω=2,

∴周期T=

2
=π;

(2)∵f(

α
2
+
π
8
)=
2
sin[2(
α
2
+
π
8
)+
π
4
]=
2
sin(α+
π
2
)=
2
cosα=
5
2

∴cosα=

10
4

又α∈[0,

π
2
],∴sinα=
1-cos2α
=
6
4

∵f(

β
2
+π)=
2
sin[2(
β
2
+π)+
π
4
]=
2
sin(β+
π
4
+2π)=
2
sin(β+
π
4
)=
2

∴sin(β+

π
4
)=1,

∵β∈[0,

π
2
],∴β+
π
4
∈[
π
4
4
],

∴β+

π
4
=
π
2
,即β=
π
4

则sin(α+β)=sin(α+

π
4
)=sinαcos
π
4
+cosαsin
π
4
=
3
+
5
4

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