寻找高手解题

证:
sinθ+cosθ=2sinα```````①
sinθcosθ=(sinβ)^2`````②

将①式平方,得 4(sinα)^2 = 1 + sin(2θ)
变形得 2 - 4(sinα)^2 = 2 - [1 + sin(2θ)]
所以得 2cos(2α) = 1 - sin(2θ)
再将②式变形,得 1 - 2sinθcosθ = 1 - 2(sinβ)^2
然后就得 1 - sin(2θ) = cos(2β)
所以就有 2cos(2α) = cos(2β) = 1 - sin(2θ) ````````③

又因为
1 - sin(2θ) = 1 - 2sinθcosθ
= (sinθ)^2 + (cosθ)^2 - 2sinθcosθ
配方:所以 1 - sin(2θ) = (sinθ - cosθ)^2
又因为 sinθ - cosθ = √2*[(√2/2)sinθ + (√2/2)cosθ]
=√2*cos(π/4 - θ) = -√2*cos(π/4 + θ)
所以(sinθ - cosθ)^2 = [-√2*cos(π/4 + θ)]^2 = 2[cos(π/4+θ)]^2
所以1 - sin(2θ) = 2[cos(π/4+θ)]^2 ````````④

综上所述: 由 ③④ 得:
2cos(2α)=cos(2β)=2[cos(π/4+θ)^2 得证!