设sinθ+cosθ=-√2,则cos2θ=_______sin4θ=_______
2sinα=sinθ+cosθ,sin²β==sinθcosθ.求证cos2β=2cos2α=2cos
若cos2θ=√2/3,则sinθ^4+cos^4的值为
cos2θ=√2/3,则cos^4-sin^4的值为
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
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
sinθ+sin2θ/1+cosθ+cos2θ=
求证cos^4θ-sin^4θ=cos2θ
2sinθ+cosθ/sinθ-3cosθ=-5,求cos2θ+4sinθ
sin2θ+sinθ/2cos2θ+2sin^θ+cosθ=tanθ 数学题
sinθ+2cosθ=0,则2cos2θ-sin2θ=