问题 解答题
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.
答案

y=sin4x+2

3
sinxcosx-cos4x

=sin4x-cos4x+2

3
sinxcosx

=(sin2x+cos2x)(sin2x-cos2x)+2

3
sinxcosx

=-cos2x+

3
sin2x

=2(sin2xcos

π
6
-cos2xsin
π
6

=2sin(2x-

π
6

∴T=

2
=π,ymin=-2,

又∵-

π
2
+2kπ≤2x-
π
6
π
2
+2kπ,

∴-

π
3
+2kπ≤2x≤
3
+2kπ,即-
π
6
+kπ≤x≤
π
3
+kπ,

所以y=2sin(2x-

π
6
)的单调增区间是[-
π
6
+kπ,
π
3
+kπ]

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