Question

cos(a+b)=1/3 cos（a-b）=1/2 则log5（tanA*tanB）=?

cos(a+b)=1/3 cos（a-b）=1/2 则log5（tanA*tanB）=? 请大家帮忙回答一下哈

Answer 1

cos(a+b)=cosacosb-sonasinb=1/3
cos(a-b)=cosacosb+sonasinb=1/2
相加，2cosacosb=5/6
cosacosb=5/12
所以sinasinb=1/12
tanatanb=sinasinb/cosacosb=1/5
所以原式=log5(1/5)
=log5(5^(-1))
=-1*log5(5)
=-1

Answer 2

解：由题意可得
cos(a+b)=cosacosb-sinasinb=1/3
cos(a-b)=cosacosb+sinasinb=1/2
所以sinasinb=1/12,cosacosb=5/12
所以sinasinb/(cosacosb)=tanatanb=1/5
所以log5（tanA*tanB）=-1