函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4)的最小正周期为______,单调减区间为______.-数学

题目简介

函数f(x)=cos(2x-π3)+2sin(x-π4)sin(x+π4)的最小正周期为______,单调减区间为______.-数学

题目详情

函数f(x)=cos(2x-
π
3
)+2sin(x-
π
4
)sin(x+
π
4
)的最小正周期为______,单调减区间为______.
题型:填空题难度:中档来源:不详

答案

函数f(x)=cos(2x-class="stub"π
3
)+2sin(x-class="stub"π
4
)sin(x+class="stub"π
4

=cos(2x-class="stub"π
3
)+2sin(x-class="stub"π
4
)sin[class="stub"π
2
+(x-class="stub"π
4
)]
=cos(2x-class="stub"π
3
)+2sin(x-class="stub"π
4
)cos(x-class="stub"π
4

=cos(2x-class="stub"π
3
)+sin(2x-class="stub"π
2

=cos(2x-class="stub"π
3
)-cos2x
=cos2xcosclass="stub"π
3
+sin2xsinclass="stub"π
3
-cos2x
=
3
2
sin2x-class="stub"1
2
cos2x
=sin(2x-class="stub"π
6
),
∵ω=2,∴T=class="stub"2π
2
=π;
由正弦函数的单调减区间为[2kπ+class="stub"π
2
,2kπ+class="stub"3π
2
],k∈Z,
得到2kπ+class="stub"π
2
≤2x-class="stub"π
6
≤2kπ+class="stub"3π
2
,k∈Z,
解得kπ+class="stub"π
3
≤x≤kπ+class="stub"5π
6
,k∈Z,
则函数f(x)的单调减区间为[kπ+class="stub"π
3
,kπ+class="stub"5π
6
],k∈Z.
故答案为:π;[kπ+class="stub"π
3
,kπ+class="stub"5π
6
],k∈Z

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