问题
解答题
求函数y=sin4x+2
|
答案
y=sin4x+2
sinxcosx-cos4x3
=sin4x-cos4x+2
sinxcosx3
=(sin2x+cos2x)(sin2x-cos2x)+2
sinxcosx3
=-cos2x+
sin2x3
=2(sin2xcos
-cos2xsinπ 6
)π 6
=2sin(2x-
)π 6
∴T=
=π,ymin=-2,2π 2
又∵-
+2kπ≤2x-π 2
≤π 6
+2kπ,π 2
∴-
+2kπ≤2x≤π 3
+2kπ,即-2π 3
+kπ≤x≤π 6
+kπ,π 3
所以y=2sin(2x-
)的单调增区间是[-π 6
+kπ,π 6
+kπ]π 3