By definition, #sin(phi)# is an ordinate (Y-coordinate) of a unit vector positioned at angle #angle phi# counterclockwise from the X-axis, while #cos(phi)# is its abscissa (X-coordinate).
Obviously, #sin^2(phi)+cos^2(phi)=1#.
Since #sin(alpha)=12/13#, #cos^2(alpha)=1-(12/13)^2=25/169# and, therefore, #cos(alpha)=+-5/13# (correspondingly to an angle in the 1st quadrant and in the second one).
Since #cos(beta)=-4/5#, #sin^2(beta)=1-(4/5)^2=9/25# and, therefore, #sin(beta)=+-3/5# (correspondingly to an angle in the second quadrant and in the third one).
So, we have the following four combinations of values for #sin# and #cos# of angles #alpha# and #beta#:
(a) #sin(alpha)=12/13#, #cos(alpha)=5/13#, #sin(beta)=3/5#, #cos(beta)=-4/5#
(b) #sin(alpha)=12/13#, #cos(alpha)=-5/13#, #sin(beta)=3/5#, #cos(beta)=-4/5#
(c) #sin(alpha)=12/13#, #cos(alpha)=5/13#, #sin(beta)=-3/5#, #cos(beta)=-4/5#
(d) #sin(alpha)=12/13#, #cos(alpha)=-5/13#, #sin(beta)=-3/5#, #cos(beta)=-4/5#
Now we can use the formula
#sin(alpha+beta)=sin(alpha)*cos(beta)+cos(alpha)*sin(beta)#
(a) #sin(alpha+beta)=12/13 * (-4/5) + 5/13 * 3/5=-33/65#
(b) #sin(alpha+beta)=12/13 * (-4/5) + (-5/13) * 3/5=-63/65#
(c) #sin(alpha+beta)=12/13 * (-4/5) + 5/13 * (-3/5)=-63/65#
(d) #sin(alpha+beta)=12/13 * (-4/5) + (-5/13) * (-3/5)=-33/65#
So, we have two solutions:
#sin(alpha+beta) = -33/65# and #sin(alpha+beta) = -63/65#.
