cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB

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cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanBcos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanBcos(A+B)=1/5,cos(A-B)=3/5,求t

cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB
cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB

cos(A+B)=1/5,cos(A-B)=3/5,求tanA*tanB
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

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