求证sin^4+cos^4=1-1/2sin^2(2α)

求证：sinα^4+cosα=1-2sinα^2cosα^2 求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 求证2cos^2a+sin^4a=cos^4a+1 求证：2cos^2θ+sin^4θ=cos^4+1 求证：2cos^2θ+sin^4θ=cos^4+1 求证sin^4+cos^4=1-1/2sin^2(2α) 求证sin^4a+cos^4a=1-2sin^2acos^2a 求证sin^4x+cos^4x=1-2sin^2xcos^2x 求证（cos^2x-sin^2x)(cos^4x+sin^4x）+1/4sin2xsin4x=cos2x 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 求证：1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3 求证3-sin^4a-cos^4a/2cos^a=1+tan^a+sin^a 速答 谢谢 求证：（1-sinα+cosα）/（1+sinα+cosα）=tan(π/4-α/2) 求证2(cosθ -sinθ )/1+sinθ +cosθ =tan(π/4-θ /2)-tanθ /2 求证：2（cos θ -sin θ )/(1+sinθ +cosθ)=tan(∏/4- θ /2)-tan(θ /2) 求证(1-2sinαcosα)/(cos²α-sin²α)=tan(π/4-α) 求证:1-2sinαcosα/cos²α-sin²α=tan（π/4-α）