求sin^4x+cos^4x=1-2sin^2xcos^2x的详细答案!
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
求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.
求函数y=2sin xcos x+2sin x+2cos x+4的值域
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/2
求函数y=sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期和值域
求函数y=sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期 值域
求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)的最小正周期、最大值和最小值
求函数f(x)=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx-1/2sinxcosx+1
化简cos^4x+sin^2xcos^2x+sin^2x
求函数fx=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx的最小正周期,最大值和最小值
求函数f(x)=(sin⁴x+cos⁴x+sin²xcos²x)/(2-si