1、f(x)=sin^2(x+π/4)-sin^2(x-π/4)的最小正周期是?2、已知sina=cos2a,π/2_数学解答

1、f(x)=sin^2(x+π/4)-sin^2(x-π/4)的最小正周期是?
2、已知sina=cos2a,π/2

人气：380 ℃　时间：2020-03-18 01:45:21

解答

1、f(x)=sin²(x+π/4)-sin²(x-π/4)=sin²[π/2+(x-π/4)]-sin²(x-π/4)=cos²（x-π/4）-sin²(x-π/4)
=cos2(x-π/4)=cos(2x-π/2)=sin2x.
∴最小正周期是π.
2.∵sina=cos2a,∴sina=1-2sin²a,解得sina=1/2或sina=-1
∵π/2