设tana=1/7,tanβ=1/3,且a,β都是锐角,求证:a+2β=π/4

问题描述:

设tana=1/7,tanβ=1/3,且a,β都是锐角,求证:a+2β=π/4

tan(2β)=1/3*2/(1-1/3*1/3)=3/4
tan(a+2β)=(1/7+3/4)/(1-1/7*3/4)
=(4+21)/(28-3)
=1
tan(2β)>0且0所以0tan(a+2β)>0且0所以0所以a+2β=π/4