Question #81c80

2 Answers
Dec 25, 2017

#A=45#
#B=15#

Explanation:

#sin(A-B)=1/2#

Therefore #A-B# = #30# because #sin 30=1/2#

#cos(A+B)=1/2#

Therefore #A+B# = #60# because #cos 60=1/2#

From there,

#A-B+A+B=30+60#

#2A=90#

#A=45#

Replace #A# with #45# in any of the two equations,

we get #B=15#.

Dec 25, 2017

Here is a plot of #color(red)(sin(A-B)=1/2)# and #color(blue)(cos(A+B)=1/2#

www.desmos.com/calculator

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)]#