函数y=cos(2x+3π)·sin(2x-π)的周期为
网友回答
解y=cos(2x+3π)·sin(2x-π)
=-cos2x*（-sin2x）
=1/2sin4x
周期T=2π/4=π/2
======以下答案可供参考======
供参考答案1:
y=cos(2x+3π)sin(2x-π)
=-cos(2x)[-sin(2x)]
=cos(2x)sin(2x)
=(1/2)sin(4x)
最小正周期Tmin=2π/4=π/2
供参考答案2:
y=cos(2x+3π)·sin(2x-π)
=(-cos2x)(-sin2x)=1/2sin4x
函数y=cos(2x+3π)·sin(2x-π)的周期为2π/4=π/2