Question

已知α为第二象限的角,cosα=-4/5,β为第一象限的角,sinβ=12/13,求tan(2α-β)的值

Answer 1

cosa=-4/5 ,sina=3/5 ,sinβ=12/13, cosβ=5/13 ,tana=-3/4
sin(a-β)=15/65+48/65=63/65 ,cos(a-β)=-20/65+36/65=16/65 ,tan(a-β)=63/16
tan(2a-β)=tan[a+(a-β)]=[tana+tan(a-β)]/[1-tanatan(a-β)]
=[-3/4+63/16]/[1-(-3/4)(63/16)]
=(51/16)/(253/64)
=204/253

追问

cos(a-β)=-20/65+36/65=16/65这里我有点不懂

追答

cos(a-β)=cosacosβ+sinasinβ
=(-4/5)(5/13)+(3/5)(12/13)
=-20/65+36/65
=16/65