题目内容
函数y=
sinxcosx-sin2x的最小正周期为______,最大值为______.
| 3 |
函数y=
sinxcosx-sin2x
=
sin2x-
=
sin2x+
cos2x-
=sin(2x+
)-
,
∵ω=2,∴T=
=π;
又-1≤sin(2x+
)≤1,即sin(2x+
)的最大值为1,
∴函数的最大值为1-
=
.
故答案为:π;
| 3 |
=
| ||
| 2 |
| 1-cos2x |
| 2 |
=
| ||
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
=sin(2x+
| π |
| 6 |
| 1 |
| 2 |
∵ω=2,∴T=
| 2π |
| 2 |
又-1≤sin(2x+
| π |
| 6 |
| π |
| 6 |
∴函数的最大值为1-
| 1 |
| 2 |
| 1 |
| 2 |
故答案为:π;
| 1 |
| 2 |
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