f(x)=3sin(ωx+π/6),ω>0,x∈R,且以π/2为最小正周期.f(α/4+π/12)=9/5,求sin(π/4+2α)

题目

f(x)=3sin(ωx+π/6),ω>0,x∈R,且以π/2为最小正周期.f(α/4+π/12)=9/5,求sin(π/4+2α)

答案

∵f(x)以π/2为最小正周期
∴2π/ω=π/2
∴ω=4
∴f(x)=3sin(4x+π/6)
∵f(α/4+π/12)=9/5
∴3sin(α+π/3+π/6)=3sin(α+π/2)=9/5
∴cosα=3/5
∴sinα=±√(1-3²/5²)=±4/5
∴sin2α=2sinαcosα=±24/25
cos2α=2cos²α-1=-7/25
∴sin(π/4+2α)=(√2/2)(sin2α+cos2α)
=(√2/2)(±24/25-7/25)
=(√2/2)(17/25)或(√2/2)(-31/25)
=(17√2)/50或(-31√2)/50