求证sin^4x+cos^4x=1-2sin^2xcos^2x

求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 求证sin^4x-cos^4x=sin^2x-cos^2x 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 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：1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) 求证 sinˇ4X+sin²Xcos²X+cos²X = 1 求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x ：求证:(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x求证:(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x我算到右边=cos2x（1-sin^2xcos^2x） 接下来该怎么办? ：求证:(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x求证:(cos^2x-sin^2x)(cos^4x+sin^4x)+1/4sin2xsin4x=cos2x我算到右边=cos2x（1-sin^2xcos^2x） 接下来该怎么办? 求证：(sin x+cos x)(1-tan x)=(2sin x)/(tan2x) 求证：4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x谢谢～～～ 已知sinθ+cosθ=2sinx,sinθcosθ=sin²y,求证：4cos²2x=cos²2y 求证 cos^8(x)-sin^8(x)=cos2x【1-1/2sin^2（2x)】