-(2cos^2x-1-2sin^2x)cos^2x 这个怎么变成 -cos^2x+4sin^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
化简cos^4x+sin^2xcos^2x+sin^2x
2cos x (sin x -cos x)+1
(sin^2x+cos^2x )(sin^2x-cos^2x)怎么得sin^2x-cos^2x?
证明sin4次方x+cos四次方x=1-2sin²xcos²x
cos 2x /sin^2 x*cos^2 x不定积分
求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
lim(sin(x^2*cos(1/x)))/x怎么做?
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/2
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x