Question

已知函数fx=x^2-4mx-3的图像经过A(tana,0),B(tanb,0)两点,求2cos2acos2b+sin2(a+

b）+2sin^2(a-b)的值

Answer 1

由题意知tana和tanb是方程x²-4mx-3=0的两个根，
∴tana+tanb=4m，tanatanb=-3
∴tan(a+b)=(tana+tanb)/(1-tanatanb)=4m/4=m
∴2cos2acos2b+sin2(a+b)+2sin²(a-b)
=2cos2acos2b+sin2(a+b)+1-cos2(a-b)
=2cos2acos2b+sin2(a+b)+1-cos2acos2b-sin2asin2b
=cos2acos2b-sin2asin2b+sin2(a+b)+1
=cos2(a+b)+sin2(a+b)+1
=[1-tan²(a+b)]/[1+tan²(a+b)]+2tan(a+b)/[1+tan²(a+b)]+1
=(1-m²)/(1+m²)+2m/(1+m²)+1
=2(m+1)/(m²+1)

Answer 2

由二次函数的性质，x1+x2=-b/a
x1*x2=c/a
则有tana+tanb=1/2
tana*tanb=-1
tan(a+b)=(tana+tanb)/(1-tana*tanb)=1/4
则tan(a+b)=sin(a+b)/cos(a+b)
可以的结论了