sin^4xcos^2x

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）原式=sin^8X-2sin^4xcos^4X+cos^8X+4sin^2Xcos^2x=（sin^4x-cos^4x）^2+4sin^2Xcos^2

积分sin^4xcos^2xdx利用三角学解以下积分(如图)Hints:sin^4xcos^2x=sin^2x(sinxcosx)^2方法1：原式=∫sin⁴x cos²x =∫sin⁴

1/(sin^4xcos^4x)积分

∫1/(sin^2xcos^2x)∫1/(sin²xcos²x)dx=∫4/(4sin²xcos²x)dx=4∫1/sin²2xdx=2∫csc²2xd(2x)=-2cot2x+C

sin^2xcos^2x不定积分∫sin^2xcos^2xdx=1/4∫(sin2x)^2dx=1/8∫(1-cos4x)dx=1/8(x-1/4sin4x)+C=x/8-(sin4x)/32+C满意请采纳,

sin^2xcos^3x的不定积分,∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C实际非常简单∫sin^2x(1-sin^2

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x+3sin^2xsin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^

求证sin^4x+cos^4x=1-2sin^2xcos^2x求证sin^4x+cos^4x=1-2sin^2xcos^2x证:sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xc

sin^4x+cos^4x=1-2sin^2Xcos^2X谢谢额sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2Xcos^2X=1-2sin^2Xcos^2X

化简sin^4x+sin^2xcos^2x+cos^2x=?(请写过程)x就不写了,原式等于sin^2(sin^2+cos^2)+cos^2=sin^2*(1)+cos^2=sin^2+cos^2=1sin^4x表示(sinx)^4还是(s

化简cos^4x+sin^2xcos^2x+sin^2xcos^4x+sin^2xcos^2x+sin^2x=cos^4x+(1-cos²x)cos²x+sin²x=cos^4x+cos²x-cos^

sin^4+sin^2xcos^2x+cos^2x证明=1sin^4+sin^2xcos^2x+cos^2x=sin^2x(sin^2x+cos^2x)+cos^2x=sin^2x+cos^2x=1.

化简(sin^4+cos^4+sin^2xcos^2x)/(2-sin2x)(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)=[(sin²x+cos²x)²-2sin²xc

化简[1-(sin^4x-sin^2xcos^2x+cos^4x)]/(sin^2x)+3sin^2x注意到sin^2x+cos^2x=1所以1=(sin^2x+cos^2x)^2=sin^4x+2sin^2xcos^2x+cos^4x1-

化简：cos^4x+sin^2xcos^2x+cos^2x原式=cos^2x(cos^2x+sin^2x+1)=2cos^2x=cos2x+1

cosxsinb-sinxcosb和sinxcosx- 行吗?

1-2sinxcosx/cos的平方x1-2sinxcosx/(cosx)^2=1-2sinx/cosx=1-2tanx1-2tanx

求函数f(x)=sin^4x+cos^4x+sin²xcos²x/2-sin2x的最小正周期,最大值和最小值.f(x)=(sin^4x+cos^4x+sin²xcos²x)/(2-sin2x)=[(s

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?那个前半括号里面相加等于一

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