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均為正數,tan[(a+b)/2]也為正數且大于1,所以a、b同象限,cos(a-b)為正
為什么推出a、b同象限就可以說明cos(a-b)為正的呢?
tan(a+b)