Question

已知0<a<π/2,π/2<B<π，且tana/2=1/2,sin(a+B)=5/13.求（1）cosa和cosB的值

（2）tan(a-B)/2的值

Answer 1

tana=2(tana/2)/[1-(tana/2)^2]
=2*1/2/[1-1/4]
=1/(3/4)
=4/3

sina=4/5,cosa=3/5

sin(a+b)
=sinacosb+cosasinb
=4/5cosb+3/5sinb
=4/5cosb+3/5√(1-cos^2 b)
5/戚搜13=4/5cosb+3/5√(1-cos^2 b)
5/13-4/5cosb=3/5√(1-cos^2 b)
25/169+16/25cos^2b-8/13cosb=9/25(1-cos^2 b)
25/169+cos^2b-8/13cosb-9/25=0
cos^2b-8/13cosb-896/4225=0
cos^2b-8/13cosb+16/169-16/169-896/4225=0
(cosb-4/13)^2-1296/4225=0
(cosb-4/13-36/65)(cosb-4/13+36/65)=0
(cosb-20/65-36/65)(cosb-20/65+36/65)=0
(cosb-56/65)(cosb+16/65)=0
cosb=56/65或cosb=-16/65
因为90°<b<180°
所以cosb=-16/65
sinb=63/65
tanb=-63/16

tanb=2(tanb/2)/[1-(tanb/2)^2]
-63/16=2(tanb/2)/[1-(tanb/2)^2]
-32(tanb/2)=63[1-(tanb/2)^2]
63(tanb/2)^2-32(tanb/2)-63=0
[9(tanb/洞改2)+7][7(tanb/2)-9]=0
tanb/2=-7/9或tanb/2=9/7
90°<b<180°
45°<b<90°
所以tanb/2=9/高颤历7

tan(a-b)/2
=(tana/2-tanb/2)/(1+tana/2*tanb/2)
=(1/2-9/7)/(1+1/2*9/7)
=(7/14-18/14)/(1+9/14)
=(-11/14)/(23/14)
=-11/23