函数f(x)=sinx(sinx-cosx)的单调递减区间是()A.[2kπ+π8,2kπ+58π](k∈Z)B.[kπ+π8,kπ+58π](k∈Z)C.[2kπ-38π,2kπ+π8](k∈Z)D

题目简介

函数f(x)=sinx(sinx-cosx)的单调递减区间是()A.[2kπ+π8,2kπ+58π](k∈Z)B.[kπ+π8,kπ+58π](k∈Z)C.[2kπ-38π,2kπ+π8](k∈Z)D

题目详情

函数f(x)=sinx(sinx-cosx)的单调递减区间是(  )
A.[2kπ+
π
8
,2kπ+
5
8
π]  (k∈Z)
B.[kπ+
π
8
,kπ+
5
8
π] (k∈Z)
C.[2kπ-
3
8
π,2kπ+
π
8
]( k∈Z)
D.[kπ-
3
8
π,kπ+
π
8
]( k∈Z)
题型:单选题难度:中档来源:不详

答案

函数f(x)=sinx(sinx-cosx)=sin2x-sinxcosx=class="stub"1-cos2x
2
-class="stub"1
2
sin2x
=-
2
2
sin(2x+class="stub"π
4
)+class="stub"1
2

-class="stub"π
2
+2kπ≤2x+class="stub"π
4
≤class="stub"π
2
+2kπ
,解得kπ-class="stub"3
8
π≤x≤kπ+class="stub"π
8
(k∈Z).
[kπ-class="stub"3
8
π,kπ+class="stub"π
8
]
(k∈Z).
故选D.

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