Question

sina=3/5,tan(pi-b)=1/2,tan(a-2b)=

Answer 1

已知sina=3/5,a∈(π/2,π),tan(π-b)=1/2,则tan(a-2b)=

tan(π-b)＝-tan(b)=1/2

tan(b)=-1/2

tan(2b)=2*tan(b)/(1-tanb^2)=-4/3

tan(-2b)=4/3

sina=3/5

tan(a)=-3/4

tan(a-2b)=[tan(a)-tan(2b)]/[1+tan(a)*tan(2b)]=7/24