Question

已知a、B都是锐角,且sinB=sinacos(a+B),当tanB取最大值时，求tan(a+B)的值

Answer 1

sinB=sinacos(a+B)
=sin[(a+B)-a]=sin(a+B)cosa-cos(a+B)sina
sin(a+B)cosa=2cos(a+B)sina
tan(a+B)=2tana
tanB=[tan(a+B)-tana]/[1+tan(a+B)tana]
=tana/[1+2(tana)^2]
[1+2(tana)^2]>=2*√[1*(2tana)^2]=2√2tana
1=√2tana时，tana=1/√2,tanB最大
tanB=1/(2√2)
tan(a+B)=(tana+tanB)/谨仿(1-tanatanB)
=(1/√2+1/(2√2))/(1-(1/√2)(1/(2√2)))
=3/(2√樱晌纯2)/(3/4)
=4/(2√脊咐2)=√2