已知sina cosa=1 3 则tan二次方a
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sina+cosa=根号2(sina+cosa)²=2sin²a+cos²a+2sinacosa=21+2sinacosa=2sinacosa=1/2
sina+cosa=1/3,故两边除以cosatana+1=1/(3cosa),tana-1/(3cosa)=-1sina+cosa=1/3,两边平方1+2sinacosa=1/9,故sinacosa
tana=sina/cosa=2则sina=2cosasina^2+cosa^2=1即4cosa*2+cosa^2=1cosa^2=1/5sina^2=4/5sina^2+sina*cosa-2cos
cosa-sina=-根号(1-2sinacosa)=-根号(1-1/4)=-√3/2答案:B-根号3/2
(sina-cosa)^2=25/16=sin^2(a)+cos^2(a)-2sin(a)cos(a)sin^2(a)+cos^2(a)=1所以2sin(a)cos(a)=-9/16sin(a)cos
解:sin^2a+cos^2a=(sina-cos)^2+2sinacoa=(sina-cosa)^2+2/5=1,(sina-cosa)^2=3/5,sina-cosa=±√15/5.
∵在△ABC中,sinAcosA=sinBcosB,∴12sin2A=12sin2B,∴sin2A=sin2B,又sin2B=sin(π-2B),∴2A=2B或2A=π-2B,∴A=B或A+B=π2,
√(1-2sinacosa)/(sina-cosa)=√(sin^2a-2sinacosa+cos^2a/(sina-cosa)=√(sina-cosa)^2/(sina-cosa)因为sina>co
sina+cosa/sina-cosa=2同时除以cosa(1+tana)/(tana-1)=2(1+tana)/(1-tana)=-2(tan45+tana)/(1-tan45*tana)=-2ta
除以(sinA^2+cosA^2)就看出答案了.
你的题目根本就没有写清楚!假设:sina-cosa=m——因为这个你没有写,所以就用字母代替了!那么:(sina-cosa)^2=m^2===>sin^2a-2sinacosa+cos^2a=m^2=
tanA=2运用公式sin2A=2tanA/(1+tan^2A)=4/5sin^2A=tan^2A/(1+tan^2A)=4/5则sinAcosA=sin2A/2=2/5(sinAcosA-sin^2
a属于(0,π/4),sina-cosa
tanA=-1/21+(2sinAcosA)/(sinA-cosA)=(sinA-cosA+2sinAcosA)/(sinA-cosA)分子分母同除以一个cosA→(tanA-1+2tanA)/tan
sinacosa=1/8所以sina,cosa>0又因为(sina)²+(cosa)²=1所以(sina)²+(cosa)²+2sinacosa=1+1/4(s
相当于原式/1=原式/(sin²a+cos²a)分子分母同除以cos^a(tan^2a+1-3tana)/tan^a+1所以原式=1/10
sin^2a-3sinacosa+cos^2a=(sin^2a-3sinacosa+cos^2a)/(sin^2a+cos^2a)=(tan^2a-3tana+1)/(tan^2a+1)=(3^2-3
(sina+cosa)/(sina-cosa)=2sina+cosa=2sina-2cosasina=3cosa令sina=3k,cosa=k所以(3k)²+k²=1k²
sin^2a+sinacosa/sin^2+1?是sin^2a+sinacosa/sin^2a+1吧?sin^2a+sinacosa/sin^2a+1=sin^2a+sinacosa/(2sin^2a