Question

已知tana=1,3sinB=sin(2a+B),求tan(a+B)的值.(要具体过程)

Answer 1

3sin(A+B-A)=sin(A+B+A)

3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA

2sin(A+B)cosA-4cos(A+B)sinA=0

sin(A+B)cosA-2cos(A+B)sinA=0

两边都除以cos(A+B)cosA

tan(A+B)-2tanA=0

tan(A+B)=2tanA

因为tanA=1

所以 tan(A+B)=2