Question

已知函数f(x)=Asin(4x+φ)(A>0,0<φ<π)在x=π/16时取得最大值2

(1)求f(x)的最小正周期
(2)求f(x)的解析式
(3)若α∈[-π/2,0],f(1/4α+π/16)=6/5,求sin(2α-π/4)的值

Answer 1

∵f(x)=Asin(4x+φ)(A>0,0<φ<π)在x=π/16时取得最大值2
∴ A=2 4*π/16+φ=π/2 φ=π/2-π/4=π/4
(1) 4T=2π 最小正周期T=π/2
(2) f(x)的解析式：f(x)=2sin(4x+π/4)) (A>0,0<φ<π)
(3) α∈[-π/2,0],f(1/4α+π/16)=2sin(α+π/2) =6/5
cosα=3/5 sinα=-4/5
∵3/5<√2/2 ∴ -π/2< α<-π/4 2α在第3象限
sin2α= -24/25 cos2α=-1/5
sin(2α-π/4)=-24/25* √2/2 -1/5*√2/2=-29√2/50