求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x

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sin^4x-cos^4x=sin^2x-cos^2x 求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) 化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x) 求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x 求证 tan（2π-X）sin（-2π-X）cos（6π-X）/ sin（X+3π/2)*cos（X+3π/2)=-tanX 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 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～～～ 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 2(sin^6x+cos^6)-3(sin^4x+cos^4x) 求证：cos²x+cos²(x+a)-2cosxcosacos(x+a)=sin³a 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^8(x)-sin^8(x)=cos2x【1-1/2sin^2（2x)】 求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(tan^2-1) = sin x + cos x