Question

已知cosa=1/7,cos(a-b)=13/14,且0<b<a<帕儿/2。(1)求tan2a的值.(2)求b的值

Answer 1

已知cosa=1/7,cos(a-b)=13/14
0＜b＜a＜π/2
则0＜a-b＜π/2

所以sina=√[1-(1/7)²]=4√3/7
sin(a-b)=√[1-(13/14)²]=3√3/14

所以tan2a=sin2a/cos2a=2sinacosa/(cos²a-sin²a)
=2*(4√3/7)*(1/7)/[(1/7)²-(4√3/7)²]
=-8√3/47

cosb=cos[a-(a-b)]=cosacos(a-b)+sinasin(a-b)
=(1/7)*(13/14)+(4√3/7)*(3√3/14)
=1/2
所以b=π/3

Answer 2

cosa=1/7 0<b<a<帕儿/2 sina=4√3/7
tana=t=sina/cosa=4√3
tan2a=2tana-(1-tan^2a)=8√3/(1-48)=-8√3/47
cos(a-b)=13/14 sin(a-b)=3√3/14
cosb=cos[a-(a-b)]
=cosacos(a-b)+sinasin(a-b)
=1/7*13/14+4√3/7*3√3/14
=(13+36)/7*14
=1/2
b=π/3

Answer 3

cosa=1/7 0<b<a<帕儿/2 sina=4√3/7
tana=t=sina/cosa=4√3
tan2a=2tana-(1-tan^2a)=8√3/(1-48)=-8√3/47
cos(a-b)=13/14 sin(a-b)=3√3/14
cosb=cos[a-(a-b)]
=cosacos(a-b)+sinasin(a-b)

b=π/3