tana,tanb是一元二次方程x^2+3x-4=0的两根,(cosa-sinbsin(a+b))/(sina+sinbcos(a+b))
问题描述:
tana,tanb是一元二次方程x^2+3x-4=0的两根,(cosa-sinbsin(a+b))/(sina+sinbcos(a+b))
答
tana,tanb是一元二次方程x^2+3x-4=0的两根 --->tana+tanb=-3,tanatanb=-4 cosa-sinbsin(a+b)=cos(a+b-b)-sinbsin(a+b)=cos(a+b)cosb+sinasin(a+b)-sinasin(a+b)=cosbcos(a+b),sina+sinbcos(a+b)=sin(a+b-b)=sin(a+b)cosb-sinbcos(a+b)+sinbcos(a+b)=sin(a+b)cosb,因此(cosa-sinbsin(a+b))/(sina+sinbcos(a+b))=cosbcos(a+b)/[sin(a+b)cosb]=1/tan(a+b)=[1-tanatanb]/[tana+tanb}=[1-(-4)]/(-3)=-5/3