已知x+y=4-2cos^22θ,x-y=4sin2θ,求证x^1/2+y^1/2=2

题目

已知x+y=4-2cos^22θ,x-y=4sin2θ,求证x^1/2+y^1/2=2

答案

证明：
2x
=(x+y)+(x-y)
=4-2(cos2θ)^2+4sin2θ
=2+2[1-(cos2θ)^2]+4sin2θ
=2+2(sin2θ)^2+4sin2θ
=2[1+2sin2θ+(sin2θ)^2]
=2(1+sin2θ)^2
∴x=(1+sin2θ)^2
∵1+sin2θ≥0
∴√x=1+sin2θ
2y
=(x+y)-(x-y)
=4-2(cos2θ)^2-4sin2θ
=2+2[1-(cos2θ)^2]-4sin2θ
=2+2(sin2θ)^2-4sin2θ
=2[1-2sin2θ+(sin2θ)^2]
=2(1-sin2θ)^2
∴y=(1-sin2θ)^2
∵1-sin2θ≥0
∴√y=1-sin2θ
∴√x+√y
=(1+sin2θ)+(1-sin2θ)
=2
证毕!