为什么sin2α-sin(π/2-2α)=2cosπ/4sin(2α-π/4)?
若2sin(π4+α)=sin θ+cos θ,2sin2β=sin 2θ,求证:sin&
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证:sin2θ+2cos2β=3
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
函数y=sinα+cosα-4sinαcosα+1,且2sin^2α+sin2α/1+tanα=k,π/4
已知sin(π-α)=-2sin(π/2+α)则cos²α+sin2α
若sin2α=sinα+cosα,则根号2*sin(α+π/4)等于多少
化简sin2α-2cos²α/sin(α-π/4)
已知sinα+cosα=三分之根号二,sinα-cosα=-4/3,且α∈(-π/2,0),计算(1+sin2α+cos
已知tan(π/4+α)=2 求(cos2α-sin2α)/(2sinαcosα+cos^2α)
若2sin(π/4+θ)=sinθ+cosθ,2sin^2(β)=sin2θ,求证sin2α+1/2cos2β=0
已知α∈﹙0,π/2),且2Sinα-SinαCosα-3Cosα=0.求[Sin(α+π/4)]/(Sin2α+Cos
sin2α=sin^2(α+π/4)-cos^2(α+π/4)=2sin^2(a+π/4)-1=1-2cos^2(α+π