已知cos(a+b)=1/5、cos(a-b)=3/5,求tana .tanb的值?
问题描述:
已知cos(a+b)=1/5、cos(a-b)=3/5,求tana .tanb的值?
答
cos(a+b)=cosacosb-sinasinb=1/5 (1) cos(a-b)=cosacosb+sinasinb=3/5 (2)
(1)+(2) 2cosacosb=4/5 (3) (1)-(2) 2sinasinb=-2/5 (4) (4)/(3) => tanatanb=-1/2
答
cos(a+b)=cosa*cosb-sina*sinb;cos(a-b)=cosa*cosb+sina*sinb;cosa*cosb=2/5;sina*sinb=1/5;故tana*tanb=1/2
答
cos(a+b)=1/5、cos(a-b)=3/5,
cosacosb-sinasinb=1/5
cosacosb+sinasinb=3/5
解得:sinasinb=1/5,cosacosb=2/5
所以tana .tanb=1/2