数学题；三角函数求证:sin^4x+cos^4x=1-2sin^2xcos^2x

sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 求证 sinˇ4X+sin²Xcos²X+cos²X = 1 化简cos^4x+sin^2xcos^2x+sin^2x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2 求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/2 求证 cos*xcos*y + sin*xsin*y + sin*xcos*y + xin*ycos*x = 1 求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域. 求函数y=2sin xcos x+2sin x+2cos x+4的值域 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 求函数f(x)=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx-1/2sinxcosx+1 证明sin4次方x+cos四次方x=1-2sin²xcos²x