求函数y=sin4x+23sinxcosx-cos4x的最小正周期、最小值和单调递增区间.-数学

题目简介

求函数y=sin4x+23sinxcosx-cos4x的最小正周期、最小值和单调递增区间.-数学

题目详情

求函数y=sin4x+2
3
sinxcosx-cos4x
的最小正周期、最小值和单调递增区间.
题型:解答题难度:中档来源:不详

答案

y=sin4x+2
3
sinxcosx-cos4x

=sin4x-cos4x+2
3
sinxcosx
=(sin2x+cos2x)(sin2x-cos2x)+2
3
sinxcosx
=-cos2x+
3
sin2x
=2(sin2xcosclass="stub"π
6
-cos2xsinclass="stub"π
6

=2sin(2x-class="stub"π
6

∴T=class="stub"2π
2
=π,ymin=-2,
又∵-class="stub"π
2
+2kπ≤2x-class="stub"π
6
class="stub"π
2
+2kπ,
∴-class="stub"π
3
+2kπ≤2x≤class="stub"2π
3
+2kπ,即-class="stub"π
6
+kπ≤x≤class="stub"π
3
+kπ,
所以y=2sin(2x-class="stub"π
6
)的单调增区间是[-class="stub"π
6
+kπ,class="stub"π
3
+kπ]

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