已知a=(2,cosx),b=(sin(x+π6),-2),函数f(x)=a•b.(1)求函数f(x)的单调增区间;(2)若f(x)=65,求cos(2x-π3)的值.-数学

题目简介

已知a=(2,cosx),b=(sin(x+π6),-2),函数f(x)=a•b.(1)求函数f(x)的单调增区间;(2)若f(x)=65,求cos(2x-π3)的值.-数学

题目详情

已知
a
=(2,cosx),
b
=(sin(x+
π
6
),-2),函数f(x)=
a
b

(1)求函数f(x)的单调增区间;
(2)若f(x)=
6
5
,求cos(2x-
π
3
)的值.
题型:解答题难度:中档来源:泰安一模

答案

(1)∵f(x)=
a
b
=2sin(x+class="stub"π
6
)-2cosx
=2sinxcosclass="stub"π
6
+2cosxsinclass="stub"π
6
-2cosx
 
=
3
sinx-cosx=2sin(x-class="stub"π
6
) …(5分)
-class="stub"π
2
+2kπ≤x-class="stub"π
6
≤class="stub"π
2
+2kπ
,k∈z,得,-class="stub"π
3
+2kπ≤x≤class="stub"2π
3
+2kπ
. …(7分)
故函数f(x)的单调增区间为[-class="stub"π
3
+2kπ ,  class="stub"2π
3
+2kπ
],k∈z.…(8分)
(2)由(1)可得f(x)=class="stub"6
5
即 sin(x-class="stub"π
6
)=class="stub"3
5
.…(10分)
∴cos(2x-class="stub"π
3
)=1-2sin2(x-class="stub"π
6
)
=class="stub"7
25
.…(12分)

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