题目内容
函数y=2sin(x+
π
解析:y=2sin(x+
)cos(x+
)=1-cos(2x+
=1+sin2x,∴T=
=π.
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相关题目
若将函数y=f(x)的图象按向量a=(
,1)平移后得到函数y=2sin(x-
)+1的图象,则函数y=f(x)单调递增区间是( )
| π |
| 6 |
| 5π |
| 6 |
A、[
| ||||
B、[
| ||||
C、[
| ||||
D、[
|
题目内容
函数y=2sin(x+
π
解析:y=2sin(x+
)cos(x+
)=1-cos(2x+
=1+sin2x,∴T=
=π.
| π |
| 6 |
| 5π |
| 6 |
A、[
| ||||
B、[
| ||||
C、[
| ||||
D、[
|