求证：cos'2θ-cos2θcos4θ=sin'23θ

求证2cos2θ+sin4θ=cos4θ+1 2sinα=sinθ+cosθ,sin²β==sinθcosθ.求证cos2β=2cos2α=2cos 求证cos^4θ-sin^4θ=cos2θ 已知sinθ+cosθ=2sinα,sinθcosθ=(sinβ)^2,求证4(cos2α)^2=(cos2β)^2 已知sinθ+cosθ=2sinα,sinθ·cosθ=sin²β,求证：2cos2α=cos2β. 三角数列题：sinθ sinα cosθ成等差数列,sinθ sinβ cosθ为等比数列,求证2COS2α=cos2β ：求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ). 已知sinθ+sin2θ=1，求3cos2θ+cos4θ-2sinθ+1的值． 求证：sin2θ+sinθ/2cos2θ+2sin^2θ+cosθ=tanθ 已知sinα与sinβ分别是sinθ与cosθ的等差中项与等比数列的中项,求证:2cos2α=cos2β=2{cos(π 若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证：sin2θ+2cos2β=3 若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.