Here is a plot of #color(red)(sin(A-B)=1/2)# and #color(blue)(cos(A+B)=1/2#
Given:
#sin(A-B) = 1/2" [1]"#
#cos(A+B) = 1/2" [2]"#
Using the corresponding inverse on each equation:
#A-B = sin^-1(1/2)" [1.1]"#
#A+B = cos^-1(1/2)" [2.1]"#
Because the inverses have two possible values, each of the two equations separates into two more equations and each repeats at integer multiples of #2pi#:
#A-B = pi/6+2n_1pi; n_1 in ZZ" [1.2a]"#
#A-B = (5pi)/6+2n_1pi; n_1 in ZZ" [1.2b]"#
#A+B = pi/3+2n_2pi; n_2 in ZZ" [2.2a]"#
#A+B = (5pi)/3+ 2n_2pi; n_2 in ZZ" [2.2b]"#
We have 4 combinations ,#(a,a), (a,b), (b,a), and (b,b)#.
Combination #(a,a)#
Add equation [1.2a] to equation [2.2a]:
#2A = pi/6+pi/3+2n_1pi+2n_2pi; n_1 in ZZ,n_2 in ZZ#
#A = pi/12+pi/6+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ#
#A = pi/4+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(a,a)]#
Subtract equation [1.2a] from equation [2.2a]:
#2B = pi/3-pi/6- 2n_1pi+2n_2pi; n_1 in ZZ,n_2 in ZZ#
#B = pi/6-pi/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ#
#B = pi/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(a,a)]#
The editor is complaining about how long my answer is getting so I shall decrease the amount of detail.
Combination #(a,b)#
#A = (11pi)/12+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(a,b)]#
#B = (3pi)/4- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(a,b)]#
Combination #(b,a)#
#A = (7pi)/12+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(b,a)]#
#B = -pi/4- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(b,a)]#
Combination #(b,b)#
#A = (5pi)/4+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(b,b)]#
#B = (5pi)/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(b,b)]#