sin^4x a^2 cos^24x b^2=1 a^2 b^2

(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

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)=

(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa

∵4sinα−2cosα5cosα+3sinα=4tanα−25+3tanα=10，∴tanα=-2．故答案为：-2

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

(sina)^2+sina=1(sina)^2=1-sina(cosa)^2=1-(sina)^2=1-1+sina=sina(cosa)^4=(sina)^2(cosa)^4+(cosa)^2=(s

平方差公式和sin²a+cos²a=1

=(cos^2(x))^2+sin^2(x)cos^2(x)+sin^2(x)=cos^2(x)(sin^2(x)+cos^2(x))+sin^2(x)=cos^2(x)+sin^2(x)=1

直接反用2倍角公式就可以了：2sinαcosα=sin2αcos2α=cos^2(α/2)-sin^2(α/2)原式=2sinπ/4cosπ/4=sinπ/2=1

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

sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin

若直线xa+yb＝1通过点M（cosα，sinα），则cosαa+sinαb＝ 1，∴bcosα+asinα=ab，∴（bcosα+asinα）2=a2b2．∵（bcosα+asinα）2≤

sin^4α+cos^2α+sin^2αcos^2α=sin^2α（sin^2α+cos^2α）+cos^2α=sin^2α+cos^2α=1,不懂可以HI我

∵sinx+cosx=√2.∴两边平方得1+sin2x=2.sin2x=1.===>cos2x=0.∴原式=（sin²x+cos²x)(sin²x-cos²x)

证明：∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co

sin^2(x)+cos^2(x+30)+sin(x)cos(x+30）=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30

由4sinα−2cosα5cosα+3sinα＝611得，4tanα−25tanα+3＝611，即7tanα=20，解得tanα=207，故答案为：207．

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

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