若sinγ/1+cosγ=4/5,则1-cosγ/2sinγ= 要过程
sinα^2+sinβ^2+sinγ^2=1,那么cosαcosβcosγ最大值等于
若cosα+cosβ+cosγ=0,sinα+sinβ+sinγ=0,则cos(α-β)=
若sin^2α+sinα=1 则cos^4α+cos^2α=
求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos
sinα+sinβ=sinγ cosα+cosβ=cosγ 证明cos(α-γ)
已知sinα+sinβ+sinγ=0,cosα+cosβ+cosγ=0.求cos(β
若sinθ+sin^2θ=1,则cos^2θ+cos^4θ+cos^6θ
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a
sin^4+cos^4=1,求sin+cos=
已知sinα+sinβ+sinγ=0,cosα+cosβ+cosγ=0.则cos(α-β)的值为______.
sina+sinβ+sinγ=0 cosa+cosβ+cosγ=0 则cos(β-γ)=
若sinα+cosα=1/2,则sinαcosα=?