Question

知tanα,tanβ是方程x^2-4x-2=0的两个实根,求cos^(α+β)+2sin(α+β)cos(α+β)-2s

Answer 1

tana+tanb=4

tana*tanb=-2

tan(a+b)=(tana+tanb)/(1-tanatanb)=4/3

tan(a+b)=sin(a+b)/cos(a+b)=4/3

sin(a+b)=4cos(a+b)/3

[sin(a+b)]^2+[cos(a+b)]^2=1

所以[sin(a+b)]^2=16/25

[cos(a+b)]^2=9/25

因为sin(a+b)/cos(a+b)=4/3〉0

所以sin(a+b)*cos(a+b)〉0

所以sin(a+b)*cos(a+b)=根号(16/25*9/25)=12/25

所以原式=9/25+2*12/25-2*16/25

=1/25