已知函数f(x)=cos(2x+π3)+sin2x-cos2x+23sinx•cosx(1)求函数f(x)的单调减区间;(2)若x∈[0,π2],求f(x)的最值;(3)若f(α)=17,2α是第一象

题目简介

已知函数f(x)=cos(2x+π3)+sin2x-cos2x+23sinx•cosx(1)求函数f(x)的单调减区间;(2)若x∈[0,π2],求f(x)的最值;(3)若f(α)=17,2α是第一象

题目详情

已知函数f(x)=cos(2x+
π
3
)+sin2x-cos2x+2
3
sinx•cosx
(1)求函数f(x)的单调减区间;       
(2)若x∈[0,
π
2
],求f(x)的最值;
(3)若f(α)=
1
7
,2α是第一象限角,求sin2α的值.
题型:解答题难度:中档来源:不详

答案

f(x)=class="stub"1
2
cos2x-
3
2
sin2x-cos2x+
3
sin2x        …(2分)
=
3
2
sin2x-class="stub"1
2
cos2x=sin(2x-class="stub"π
6
)                            …(3分)
(1)令class="stub"π
2
+2kπ≤2x-class="stub"π
6
class="stub"3π
2
+2kπ,解得class="stub"π
3
+kπ≤x≤class="stub"5π
6
+kπ          …(5分)
∴f(x)的减区间是[class="stub"π
3
+kπ,class="stub"5π
6
+kπ](k∈Z)                    …(6分)
(2)∵x∈[0,class="stub"π
2
],∴2x-class="stub"π
6
∈[-class="stub"π
6
class="stub"5π
6
],…(7分)
∴当2x-class="stub"π
6
=-class="stub"π
6
,即x=0时,f(x)min=-class="stub"1
2
,…(8分)
当2x-class="stub"π
6
=class="stub"π
2
,即x=class="stub"π
3
时,f(x)max=1                      …(9分)
(3)f(α)=sin(2α-class="stub"π
6
)=class="stub"1
7
,2α是第一象限角,即2kπ<2α<class="stub"π
2
+2kπ
∴2kπ-class="stub"π
6
<2α-class="stub"π
6
class="stub"π
3
+2kπ,∴cos(2α-class="stub"π
6
)=
4
3
7
,…(11分)
∴sin2α=sin[(2α-class="stub"π
6
)+class="stub"π
6
]=sin(2α-class="stub"π
6
)•cosclass="stub"π
6
+cos(2α-class="stub"π
6
)•sinclass="stub"π
6
   …(12分)
=class="stub"1
7
×
3
2
+
4
3
7
×class="stub"1
2
=
5
3
14
                              …(14分)

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