Question

已知tana,tanb是方程x^2－4x－2=0的两实根,求cos^2(a+b)+2sin(a+b)cos(a+b)－2

Answer 1

本题用到的公式:

1.一元二次方程根与系数的关系:x1+x2=-b/a x1x2=c/a

2.二倍角公式:sin2A=2sinAcosA =2tanA/(1+tanA^2)

cos2A=cos^A-sin^A=1-2*sin^A=(1-tan^A)/(1+tan^A)

tan2A=2tanA/(1-tan^A)

∵

tana,tanb是方程x^2－4x－2=0的两实根

∴

tana+tanb=4

tana*tanb=-2

∴

tan(a+b)=(tana+tanb)/[1-(tana*tanb)]=4/3

cos^2(a+b)+2sin(a+b)cos(a+b)－2sin^2(a+b)

=cos[2(a+b)]+sin[2(a+b)]-sin^2(a+b)

=cos[2(a+b)]+sin[2(a+b)]-1/2+(1/2)*cos[2(a+b)]

=(3/2)*cos[2(a+b)]+sin[2(a+b)]-1/2

(3/2)*{[1-tan^(a+b)]/[1+tan^(a+b)]}+2*tan(a+b)/[1+tan^(a+b)]-1/2

=1/25