sin^4 cos^4

(3sinα-2cosα)/(5sinα+4cosα)(分子分母同时除以cosα)=(3tanα-2)/(5tanα+4)=(-2*3-2)/(-2*5+4)=-8/(-6)=4/3sin^2α+2s

再问：再问：在你答题的时候我蛋疼做了一遍，结果好像不一样……再问：不过还是辛苦施主了

(sina)^6+(cosa)^6=(sin^a+cos^a)*[(sina)^4+(cosa)^4-sin^a*cos^a]=(sina)^4+(cosa)^4-sin^a*cos^a=[(sin^

sinα-cosα=m1-2sinacosa=m^22sinacosa=1-m^2sin^4α+cos^4α=(sin^2α+cos^2α)-2sin^2acos^2a=1-1/2(2sinacosa

证明：输入过于麻烦,用换元法吧设A=sin²A,B=sin²B∵sin^4a/sin^2b+cos^4a/cos^2b=1即A²/B+(1-A)²/(1-B)=

最后结果是1.5吧?1=(sin^2a+cos^2a)^2=sin^4+2sin^2a×cos^2a+cos^4a可以明白?1-sin^6a-cos^6asin^4a+2sin^2a×cos^2a+c

前面sinα-cosα×tanα=0后面-sin^4α-sin^2α×cos^2α-cos^2α=-sin⁴α-sin²α(1-sin²α)-cos²α=-s

d/dxsin^8(cos(4x))=8sin^7(cos(4x)dsin(cos(4x)/dx=8sin^7(cos(4x)*cos(cos(4x))dcos(4x)/dx=8sin^7(cos(4

1－（sina）^4－（cosa）^4＝［1－（sina）^2］［1＋（sina）^2］－（cosa）^4＝（cosa）^2［1＋（sina）^2］－（cosa）^4＝（cosa）^2［1＋（sina

√[(sinx)^4+4(cosx)^2]-√[(cosx)^4+4(sinx)^2]=√[((sinx)^2-2)^2]-√[((cosx)^2-2)^2]=(sinx)^2-2-[(cosx)^2

cosπ/4cosπ/3-sinπ/4sinπ/3=cos45°cos60°-sin45°sin60°=(√2/2)*(1/2)-(√2/2)*(√3/2)=(√2-√6)/4

得2/3.分母将1化成（sin^2a+cos^2a）^3,分子中的1化成（sin^2a+cos^2a）^2,得到式子等于(2sin^2acos^2a)/(3sin^2acos^2a(sin^2a+co

（sinx）^4+（cosx）^4=1即(sinx)^4+(cosx)^4+2(sinx)^2(cosx)^2-2(sinx)^2(cosx)^2=1即[(sinx)^2+(cosx)^2]^2-2(

令A=arccos(-1/3)B=arcsin(-1/4)那么有cosA=-1/3sinB=-1/4由arccos,arcsin的取值范围可知,A为钝角（90°-180°之间）B为一个负的锐角（-90

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

3/5,sin平方加cos平方等于1,cos=±3/5再问：等于多少度呢再答：arccos(±3/5),等不到整数度数再问：哦

设含α的直角三角形边长为a,b,csinα=a/ccosα=b/c(a+b)/c=4/3c=3*(a+b)/4c^2=9/16*(a+b)^2a^2+b^2=c^2结合上面得7*(a^2+b^2)=1

原式=(sin²a+cos²a)(sin²a-cos²a)=(sina+cosa)(sina-cosa)或者=(sin²a+cos²a)(s

cos^4θ-sin^4θ=(cos^2θ+sin^2θ)(cos^2θ-sin^2θ)=1*cos2θ=cos2θ

∵sinα=2cosα，∴tanα=2，∴sinα−4cosα5sinα+2cosα=tanα−45tanα+2=-16；sin2α+2sinαcosα=sin2α+2sinαcosαsin2α+co