2sin^4x+2cos^4x-cos^2x-3的最小正周期.

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sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x 证明sin^4x-cos^4x=sin^2x-cos^2x 求证sin^4x-cos^4x=sin^2x-cos^2x 证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4 用cos x表示sin^4 x -sin^2 x +cos^2 x 用cos x表示sin^4x-sin^2 x+cos^2 x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 2(sin^6x+cos^6)-3(sin^4x+cos^4x) 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 关于三角函数的一道题（高中）cos(10)*cos(20)*cos(40)=?类似cos(X)*cos(2x)*cos(4x)=?sin(x)*sin(2x)*sin(4x)=?的题目有没有什么规律 求函数y=7-4sin(x)cos(x)+4cos^2(x)-4cos^4(x)值 cos 2x(1-2sin^2 2x) + cos 4x (1+cos 2x)= 关于sin^2x+cos^2x 的应用式子：cos^6x+sin^6x=（cos^2x+sin^2x)(cos^4x-cos^2xsin^x+sin^4x)=(cos^2x+sin^2x)^2-3sin^2xcos^2x 求这两步算法中的具体思路 ps:cos^2x=cosx的二次方 求解三角函数 cos 2 X (1 - 2 sin ^2 2 X) + cos 4 X ( 1 + cos 2 X)cos 2 X (1 - 2 sin ^2 2 X) + cos 4 X ( 1 + cos 2 X) = cos 2 X cos4X (1 + cos 2 X) = cos 4 X (1 +2cos 2X) = 0 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x