已知sina cosa=1 √2求下列各式的值1tan2a 1 tan2a

(sinA+cosA)^2=1/4sin^2A+cos^2A+2sinAcosA=1/42sinAcosA=-3/4sinAcosA=-3/8sinAcosA/(sin^2A+cos^2A)=-3/8

sina+cosa=根号2（sina+cosa）²=2sin²a+cos²a+2sinacosa=21+2sinacosa=2sinacosa=1/2

sina+cosa=-1/5两边平方得sin²a+2sinacosa+cos²a=1/251+2sinacosa=1/25∴sinacosa=-12/25(1/sina)+(1/c

tan(π/4+a)=[tanπ/4+tana)/[1-tanπ/4*tana]=2(1+tana)/(1-tana)=2tana=1/31/(2sinacosa+cos²a)=(sin&#

tana=3,得sina/cosa=3sina=3cosa因sin²a+cos²a=1所以有10cos²a=1cos²a=1/101+sinacosa-cos&

sinA^2+cosA^2=11/(sinA^2-sinAcosA-2cosA^2)=(sinA^2+cosA^2)/(sinA^2-sinAcosA-2cosA^2)=(tanA^2+1)/(tan

sins2a+2sinacosa+cos2a=1/52sinacosa=-4/5sinacosa=-2/5再答：sina-cosa=sin2a-2sinacosa+cos2a=9/5

sinA+cosA=1.5平方得(sinA)^2+2sinAcosA+(cosA)^2=2.252sinAcosA=1.25=5/4sinAcosA=5/8其实,这道题是有问题的sinA+cosA的最

把前一个式子两边同时平方,（sina-cosa）^2=(-1/5)^2,得到(sina)^2+(cosa)^2-2sina*cosa=1/25由于(sina)^2+(cosa)^2=1所以sina*c

sina-cosa=根号6/2(sina-cosa)^2=3/21-2sinacosa=3/22sinacosa=-1/2sinacosa=-1/4

tana＝sina/cosa＝1/2,所以cosa＝2sina,所以根据sin^2a＋cos^2a＝1,求出sin,cos

tana=2/3万能公式sin2a＝2tana/(1+tan^2a)1/sinacosa=2/2sinacosa=2/sin2a=2(1+tan^2a)/2tana=(1+tan^2a)/tana=(

tana/（tana-1）=-1,tana=1/2sin²a+sinacosa+2=(sin²a+sinacosa)/1+2=(sin²a+sinacosa)/(sin^

sin²a+sinacosa+2=（sin²a+sinacosa）/(sin²a+cos²a)+2=(tan²a+tana)/(tan²a+

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

2sin2a=1-cos2asinacosa=1/2sin2a总=1-1/2sin2a然后用1+tan2=1/cos2得cosa=1/5sina=五分之根号二十四sin2a=25分之根号24总=1-2

左右同乘sinacosa,两边平方,令t=sinacosa,解一元二次方程答案为t=1-√2t=1+√2不符合,因为t=sinacosa=sina+cosa

用万能公式,cos2A=sin(π/2+2A)=2tan[(π/4)+A]/{1+tan^2[(π/4)+A]}=4/5sin2A=-cos[(π/2)+2A]=-{1-tan^2[(π/4)+A]}

cosa=3/2sina原式=sina^2-2sina*3/2sina+9/4sina^2=1/4sina^2又cosa^2+sina^2=1得sina^2=4/13则原式=1/13

1-cos2a/sinacosa=1-2(cos2a/2sinacosa)=1-2(cos2a/sin2a)=1-2cot2atana=0.5,tan2a=2tana/(1-tana*tana)=4/