求证；1-2sin Xcos X/cos2 X-sin2 X=1-tan X/1+tan X

求证sin^4x+cos^4x=1-2sin^2xcos^2x 求证 cos*xcos*y + sin*xsin*y + sin*xcos*y + xin*ycos*x = 1cos*xcos*y + sin*xsin*y + sin*xcos*y + sin*ycos*x = 1注意：[*] 的意思是 [ ^2 ]写下左右过程.. 求证 sinˇ4X+sin²Xcos²X+cos²X = 1 1-2sin xcos x/cos的平方x sin^4x+cos^4x=1-2sin^2Xcos^2X谢谢额 1-(sin^6x+cos^6x)=3sin^2xcos^2x 如何证明? (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x sin^4+sin^2xcos^2x+cos^2x 证明 =1 cos xsin b-sin xcos b和sin xcos x- ：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? 证明sin4次方x+cos四次方x=1-2sin²xcos²x sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简） 判别下列函数是否是周期函数?若是则求证出其周期．1)sin(1-2x);2)sin^2x;3)3-|sin4x|;4)xcos x;5)2cos (x/2) - 3sin (x/3). ∫1/(sin^2xcos^2x) 求sin^4x+cos^4x=1-2sin^2xcos^2x的详细答案! 求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.