Question

If sin(A+B)=1 and cos (A-B)=√3/2 , then find A and B.

Answer 1

sin(A+B) = 1
A+B = sin inverse of 1
A+B = π/2
cos(A-B) = √3/2
A-B = cos inverse of √3/2
A-B = π/6
A+B+A-B = π/2 + π/6
2A = 4π/6
⇒A = π/3
A+B-A+B = π/2 - π/6
2B = π/3
⇒B = π/6

Answer 2

Let us look at angles from 0 to 360 or 2π degrees.

sin (A+B)  = 1  =>  A+B =  90 deg or 2nπ + 90  or 360+90 = 450

Cos (A-B) = √3/2  =>  A - B  =  30 or  330  (-30 deg)  or 2nπ +- 30

Let A+B = 90  A-B = 30  =>    A = 60    B=30
Let A+B = 90  A-B = 330  =>  A = 210   B = -120,  same as, 240 deg

Let A+B = 450  A-B = 30  =>  A = 240   B = 210
Let A+B = 450  A-B  = 330  => A = 390 or 30  B =  60