高一数学
已知TANA=1/2,TAN(A-B)=-2/5,那么TAN(B-2A)=?
tanA=1/2 tan2A =2tanA/(1-tan^2A) =1/(1-1/4) =4/3 tan(A-B)=-2/5 =(tanA-tanB)/(1+tanAtanB) =(1/2-tanB)/(1+tanB/2) =-2/5 9/10=4tanB/5 tanB=9/8 tan(B-2A) =(tanB-tan2A)/(1+tanBtan2A) =(9/8-4/3)/(1+9/8*3/4) =(-5/24)/(60/24) =-1/12
TAN(A-B)=-2/5,可知TANB=TANA+2/5=9/10, TAN(B-2A)=9/10-1=-1/10
答:1. tan2a = tg[(a+b)+(a-b)] = [tg(a+b)+tg(a-b)]/[1-tg(a+b)*tg(a-b)] = -4/7 tan2b...详情>>