cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB

问题描述:

cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB

tanA*tanB
=[cos(A-B)-cos(A+B)]/[cos(A+B)+cos(A-B)]
=(3/5-1/5)/(1/5+3/5)
=1/2

cosAcosB=[cos(A+B)+cos(A-B)]/2=2/5
sinAsinB=[cos(A-B)-cos(A+B)]/2=1/5
所以
tanAtanB=sinAsinB/(cosAcosB)=1/2