Question

求函数y=（sinx+cosx）2+2cos2x的最小正周期=______

求函数y=（sinx+cosx）2+2cos2x的最小正周期=______．

Answer 1

y
=
(sinx
+
cosx)²
=
sin²x
+
cos²x
+
2sinxcosx
=
1
+
sin2x
y的最小正周期为:t
=
2π/2
=
π

Answer 2

y=1+sin 2x+2cos2x=sin 2x+cos 2x+2
=
2
(
2
2
sin2x+
2
2
cos2x)+2
=
2
sin（2x+
π
4
）+2．
故
最小正周期
=
2π
2
=π．
故答案为：π．