已知tana,tanb是方程3x2+5x-7=0的两根,求值
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根据二元一次方程根与系数的关系,tana+tanb= -5/3 ……(1) tana*tanb= -7/3……(2)则1 tan(a+b)=(tana+tanb)/(1-tana*tanb)= (-5/3)/(10/3)= -1/23. 上式两边平方得,tan^2 (a+b)=1/4于是1+tan^2 (a+b)=1+1/4=5/4即sin^2 (a+b)/sin^2 (a+b)+sin^2 (a+b)/cos^2 (a+b)=5/4左边合并得,[sin^2 (a+b)+cos^2 (a+b)]/cos^2 (a+b)=5/41/cos^2 (a+b)=5/4 cos^2 (a+b)=4/52 由上式得,sin^2 (a+b)=1-cos^2 (a+b)=1-4/5=1/5(1)式两边平方得,tan^2 a+tan^2 b+2tana*tanb=25/9 [^2指平方]结合(2)式,两边同减去4tana*tanb得tan^2 a+tan^2 b-2tana*tanb=25/9+28/3=25/9+84/9=109/9即(tana-tanb)^2=109/9 tana-tanb= √109/3于是,tan(a-b)=(tana-tanb)/(1+tana*tanb)= (√109/3)/(-4/3)= -√109/4按第3题的方法可得,1/cos^2 (a-b)=1+tan^2 (a-b)=1+109/16=125/16于是cos^2 (a-b)=16/125 sin^2 (a-b)=1-16/125=109/125于是sin^2 (a+b)/sin^2 (a-b)=(1/5)/(109/125)=(1/5)*(125/109)=25/109tan(a+b)和tan(a-b)皆为负,则不管a+b和a-b在第二或第四象限,sin（α+β）和sin（α-β）皆为同号,于是sin（α+β）/sin（α-β）=5/√109=5√109/109

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