已知tan(A+B)=3,tan(A-B)=5，求tan2A,tan2B的值

Answer 1

(tanA+tanB)/(1-tanAtanB)=3
(tanA-tanB)/(1+tanAtanB)=5
解得tanA=(7±√65)/4
tanB=8±√65
tan2A=2tanA/(1-tanA^2)=±4/7
tan2B=2tanB/(1-tanB^2)=±1/8
好复杂，仔细算清楚，能输入字太少

Answer 2

tan2A=tan[(A+B)+(A-B)]
=[tan[(A+B)+tan(A-B)]/[1-tan[(A+B)*tan(A-B)]
=(3+5)/(1-3*5)=-4/7

tan2B=tan[(A+B)-(A-B)]
=[tan[(A+B)-tan(A-B)]/[1+tan[(A+B)*tan(A-B)]
=(3-5)/(1+3*5)=-1/8

Answer 3

tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan2a=tan[(a+b)+(a-b)]=(3+5)/(1-3*5)=-4/7
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
tan2b=tan[(a+b)-(a-b)]=(3-5)/(1+3*5)=-1/8

Answer 4

tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan2a=tan[(a+b)+(a-b)]=(3+5)/(1-3*5)=-4/7
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
tan2b=tan[(a+b)-(a-b)]=(3-5)/(1+3*5)=-1/8

Answer 5

用tan的和角公式
第一个是-4/7
第二个是-1/8