作业帮·已知tan[(a+b)/2]＝(√6)/2,tanatanb=13/7,求cos(a-b)=____.
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![·已知tan[(a+b)/2]＝(√6)/2,tanatanb=13/7,求cos(a-b)=____.](/uploads/image/z/6865726-22-6.jpg?t=%C2%B7%E5%B7%B2%E7%9F%A5tan%5B%28a%2Bb%29%2F2%5D%EF%BC%9D%28%E2%88%9A6%29%2F2%2Ctanatanb%3D13%2F7%2C%E6%B1%82cos%28a-b%29%3D____.)
·已知tan[(a+b)/2]＝(√6)/2,tanatanb=13/7,求cos(a-b)=____.
·已知tan[(a+b)/2]＝(√6)/2,tanatanb=13/7,求cos(a-b)=____.
·已知tan[(a+b)/2]＝(√6)/2,tanatanb=13/7,求cos(a-b)=____.
tan(a+b)
=2tan[(a+b)/2]/{1-{tan[(a+b)/2]}^2}
=√6/(1-6/4)
=-2√6
tana+tanb
=tan(a+b)*(1-tanatanb)
=-2√6*(1-13/7)
=(12√6)/7
(tana-tanb)^2
=(tana+tanb)^2-4tanatanb
=864/49-52/7
=500/49
tan(a-b)=(tana-tanb)/(1+tanatanb)
[tan(a-b)]^2
=(tana-tanb)^2/(1+tanatanb)^2
=(500/49)/(1+13/7)^2
=5/4
1+[tan(a-b)]^2=1/[cos(a-b)]^2
1+5/4=1/[cos(a-b)]^2
cos(a-b)=2/3
tana和tanb均为正数,tan[(a+b)/2]也为正数且大于1,所以a、b同象限,cos(a-b)为正
tan(a+b) =( 2tan(a+b /2) )/ (1- (tan(a+b /2))^2 )
=√6 / (1-3/2) =-2√6
(tan(a+b))^2 = 24= sec(a+b)^2 -1
sec(a+b)=5 (-5带入后,最后cos(a-b)>1无效)
cos(a+b)=1/5
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tan(a+b) =( 2tan(a+b /2) )/ (1- (tan(a+b /2))^2 )
=√6 / (1-3/2) =-2√6
(tan(a+b))^2 = 24= sec(a+b)^2 -1
sec(a+b)=5 (-5带入后,最后cos(a-b)>1无效)
cos(a+b)=1/5
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cos(a+b)=cosacosb-sinasinb
cosacosb=cos(a+b)+sinasinb=1/5+sinasinb
tanatanb=13/7=sinasinb/cosacosb
sinasinb/(1/5+sinasinb)=13/7
sinasinb=-13/25
cos(a-b)=cos(a+b)+2sinasinb=1/5+2(-13/25)
=-21/25
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