作业帮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
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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
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
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
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
1.原式=2COS（2X+π/2）-2COS（2X-π/2）=-2SIN（2X）-2SIN（2X）=-4SIN（2X）T=π
2。A=120 tanA=√3/3
3。T=π
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