Question

若aε(0,pai/2)bε(0,pai/2)且cosa=3/5tan(a-b)=-3求(1)sin(a-pai/3)

Answer 1

(1) cosa=3/5 sina=4/5

sin(a-pai/3)=sinacospai/3-cosasinpai/3=4/5*1/2-3/5*根号3/2=(4-3根号3)/10

(2) tana=4/3

tanb=tan[a-(a-b)]=[tana-tan(a-b)]/[1+tanatan(a-b)]=(4/3+3)/(1-4)=-4/9-1=-13/9