函数y=sin2x-cos2x的最小正周期为___.
问题描述:
函数y=sin2x-cos2x的最小正周期为___.
答
y=sin2x-cos2x
=√2(√2/2*sin2x-√2/2cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
所以T=2π/2=π