Question

已知cosa=1/7cos(a-b)=13/14,且0<b<a<π/2

求tan2a的值，求b~先谢谢啦·

Answer 1

cos(a-b-a)=cosb=cos(a-b)cosa+sin(a-b)sina
0<b<a<π/2,0<a-b<π/2
sina=4√3/7 sin(a-b)=3√3/14
cos(a-b-a)=cosb=cos(a-b)cosa+sin(a-b)sina
=13/14 * 1/7 +4√3/7 * 3√3/14
=13/98 + 36/98
=49/98
=1/2
b=π/3
tana=sina/cosa = 4√3
tan2a=2tana/[1-(tana)^2]=-8√3/47

Answer 2

由求sina 再求tana然后求tan2a 再将等式分开 求b