sin^4+sin^2xcos^2x+cos^2x 证明 =1

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简） ∫1/(sin^2xcos^2x) sin^2xcos^2x不定积分 sin^2xcos^3x的不定积分, (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 求证sin^4x+cos^4x=1-2sin^2xcos^2x sin^4x+cos^4x=1-2sin^2Xcos^2X谢谢额 化简sin^4x+sin^2xcos^2x+cos^2x=?(请写过程) 化简cos^4x+sin^2xcos^2x+sin^2x sin^4+sin^2xcos^2x+cos^2x 证明 =1 积分 sin^4xcos^2x dx利用三角学解以下积分(如图)Hints:sin^4xcos^2x = sin^2x(sinxcosx)^2 化简(sin^4+cos^4+sin^2xcos^2x)/(2-sin2x) sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos^2x+cos^4x 是根据什么化简的? 1-2sin xcos x/cos的平方x 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.. 求不定积分(1/sin^2xcos^2x)dx 求不定积分,∫sin^2xcos^2x dx