tanA/2tanB/2=1/3

Question

A+B≠0,sinA+sinB=2sin(A+B)

Answer 1

(1) sinA+sinB=2sin((A+B)/2)cos((A-B)/2) (trig identity).

sin(A+B)=2sin((A+B)/2)cos((A+B)/2) (trig identity).

If sinA+sinB=2sin(A+B), then 2sin((A+B)/2)cos((A-B)/2)=4sin((A+B)/2)cos((A+B)/2).

Since A+B≠0 we can remove the common factor: cos((A-B)/2)=2cos((A+B)/2).

Note that if A=B, this becomes 1=2cosA and A=π/3=B.

If we put A=B=π/3 in the original equation we get: 2sin(π/3)=2sin(π/3) showing that A=B=π/3 is a particular solution.

However, if we put A=B=π/4, for example, 2sin(π/4)≠2sin(π/2) because √2≠2. This proves that sinA+sinB=2sin(A+B) is not an identity but an equation with particular solutions.

(2) tan(A/2)/tan(B/2)=1/3.

3tan(A/2)=tan(B/2) is satisfied when 3/√3=√3 that is, when A/2=π/6 and B/2=π/3, so A=π/3 and B=2π/3.