Question

已知阿尔法,角贝塔均为锐角,且sina=5分之3,tan(a-b)=-3分之1 求sin(a-b)的值，求cosB的值，谢谢

Answer 1

a,b均为锐角，说明sina,sinb,cosa,cosb都大于0
tan(a-b)=sin(a-b)/cos(a-b)=sinacosb-sinbcosa/cosacosb+sinasinb
sina=3/5,sinb=4/5,在根据sinb*sinb+cosb*cosb=1，
就能求出cosb

Answer 2

解
1 sina=3/5, cosa=4/5. tana=3/4

2 tanb

=tan[a-(a-b)]
=[tana-tan(a-b)]/[1+tanatan(a-b)]
=[(3/4)+(1/3)]/[1-(1/4)]
=13/9
tanb=13/9
∴cosb=13/(5√10)
sin(a-b)=sinacosb-cosasinb不好打字，自己算吧！

祝您成功~