[asin(派/5)+bcos(派/5)]/[acos(派/5)-bsin(派/5)]=tan(8派/15)求b/a
热心网友
[asin(派/5)+bcos(派/5)]/[acos(派/5)-bsin(派/5)]=tan(8派/15)== [(a/b)*tan(派/5)+1]/[(a/b)-tan(派/5)]=tan(8派/15)== a/b = -[1+tan(派/5)tan(8派/15)]/[tan(派/5)-tan(8派/15)]== a/b = -1/tan[(派/5)-(8派/15)] = -1/tan(-派/3) = genhao3== b/a = (genhao3)/3