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,

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,

Answer 1

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~