题目内容
函数y=sin2x-sin4x的最小正周期T=______.
∵y=sin2x-sin4x
=sin2x(1-sin2x)
=sin2x•cos2x
=
sin22x
=
,
∴其最小正周期T=
=
.
故答案为:
.
=sin2x(1-sin2x)
=sin2x•cos2x
=
| 1 |
| 4 |
=
| 1-cos4x |
| 8 |
∴其最小正周期T=
| 2π |
| 4 |
| π |
| 2 |
故答案为:
| π |
| 2 |
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