题目内容
已知向量
=(2sin
,2cos
),
=(cos
,sin
),函数f(x)=
•
(x∈R).
(1)求函数y=f(x)的表达式;
(2)设α,β∈[0,
],f(3α+π)=
,f(3β+
)=-
,求cos(α+β)的值.
| a |
| x |
| 3 |
| x |
| 3 |
| b |
| π |
| 6 |
| π |
| 6 |
| a |
| b |
(1)求函数y=f(x)的表达式;
(2)设α,β∈[0,
| π |
| 2 |
| 16 |
| 5 |
| 5π |
| 2 |
| 20 |
| 13 |
(1)依题意得f(x)=2sin
cos
+2cos
sin
=2sin(
+
)(4分)
(2)由f(3α+π)=
得4sin[
(3α+π)+
]=
,
即4sin(α+
)=
,
∴cosα=
,
又∵α∈[0,
],
∴sinα=
=
,(8分)
由f(3β+
)=-
,
得4sin[
(3β+
)+
]=-
,即4sin(β+π)=-
,
∴sinβ=
,又∵β∈[0,
],
∴cosβ=
=
,(12分)
∴cos(α+β)=cosαcosβ-sinαsinβ=
×
-
×
=
(14分)
| x |
| 3 |
| π |
| 6 |
| x |
| 3 |
| π |
| 6 |
| x |
| 3 |
| π |
| 6 |
(2)由f(3α+π)=
| 16 |
| 5 |
| 1 |
| 3 |
| π |
| 6 |
| 16 |
| 5 |
即4sin(α+
| π |
| 2 |
| 16 |
| 5 |
∴cosα=
| 4 |
| 5 |
又∵α∈[0,
| π |
| 2 |
∴sinα=
| 1-cos2α |
| 3 |
| 5 |
由f(3β+
| 5π |
| 2 |
| 20 |
| 13 |
得4sin[
| 1 |
| 3 |
| 5π |
| 2 |
| π |
| 6 |
| 20 |
| 13 |
| 20 |
| 13 |
∴sinβ=
| 5 |
| 13 |
| π |
| 2 |
∴cosβ=
| 1-sin2β |
| 12 |
| 13 |
∴cos(α+β)=cosαcosβ-sinαsinβ=
| 4 |
| 5 |
| 12 |
| 13 |
| 3 |
| 5 |
| 5 |
| 13 |
| 33 |
| 65 |
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