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