已知函数y=sin(πX+φ)-2cos(πX+φ)(o<φ<π)的图像关于直线x=1对称,则sin2φ=?
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解由函数y=sin(πX+φ)-2cos(πX+φ)
=√5(1/√5sin(πX+φ)-2/√5cos(πX+φ))
=√5sin(πX+φ-θ)(cosθ=1/√5,sinθ=2/√5)
由函数图像关于直线x=1对称
则x=1时√5sin(π×1+φ-θ)=±√5
即sin(π+φ-θ)=±1
即sin(φ-θ)=±1
即φ-θ=kπ+π/2,k属于Z
故φ=kπ±π/2+θ,k属于Z
故sin2φ
=2sinφcosφ
=2sin(kπ±π/2+θ)cos(kπ±π/2+θ)
=2sin(±π/2+θ)cos(±π/2+θ)
=2sin(±(π/2-θ))cos(±(π/2-θ))
=-2sin(π/2-θ)cos(π/2-θ)
=-2cosθsinθ
=-2×1/√5×2/√5
=-4/5
=√5(1/√5sin(πX+φ)-2/√5cos(πX+φ))
=√5sin(πX+φ-θ)(cosθ=1/√5,sinθ=2/√5)
由函数图像关于直线x=1对称
则x=1时√5sin(π×1+φ-θ)=±√5
即sin(π+φ-θ)=±1
即sin(φ-θ)=±1
即φ-θ=kπ+π/2,k属于Z
故φ=kπ±π/2+θ,k属于Z
故sin2φ
=2sinφcosφ
=2sin(kπ±π/2+θ)cos(kπ±π/2+θ)
=2sin(±π/2+θ)cos(±π/2+θ)
=2sin(±(π/2-θ))cos(±(π/2-θ))
=-2sin(π/2-θ)cos(π/2-θ)
=-2cosθsinθ
=-2×1/√5×2/√5
=-4/5
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