求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.

化简sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x 求∫1/(sin^4x+cos^4x)dx, sin^4x+cos^4x怎么求导数? 求4sin(3x)cos(x)的导数式如题 求微积分 ∫sin^2(x)cos^4(x) dx 化简cos^4x+sin^4x 化简:cos^4(X)-sin^4(X) 化简sin^4x+cos^2x 化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x) 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 求(1-sin^6x-cos^6x)/(1-sin^4x-cos^4x)的值 求1-cos^6x-sin^6x/1-cos^4x-sin^4x的值, [（sin x-cos x）/(sin x+cos x)]^4 求定积分 积分区间0-π/4 求证明.关于cos和sin 题目有点长0 0【cos(2x) + cos(4x) + cos(6x)】 / 【sin(2x) + sin(4x) + sin(6x) 】 = cot(4x) (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（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x