use your sum-to-product formulas.
sinA + sinB = 2 sin (A+B)/2 cos (A-B)/2
now use the formula for sum of cosines, and recall that
tanθ = sinθ / cosθ
if sinA+sinB=x and cosA-cosB=y then find tan[(A-B)/2]
2 answers
Two of the lesser known identities are:
sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2] and
cosA - cosB = -2sin[A+B)/2]sin[A-B)/2]
so
x/y = ( 2sin[(A+B)/2]cos[(A-B)/2] ) / (-2sin[A+B)/2]sin[A-B)/2] )
x/y = - cos[(A-B)/2] / sin[A-B)/2]
x/y = - cot[(A-B)/2]
tan[(A-B)/2] = -y/x
sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2] and
cosA - cosB = -2sin[A+B)/2]sin[A-B)/2]
so
x/y = ( 2sin[(A+B)/2]cos[(A-B)/2] ) / (-2sin[A+B)/2]sin[A-B)/2] )
x/y = - cos[(A-B)/2] / sin[A-B)/2]
x/y = - cot[(A-B)/2]
tan[(A-B)/2] = -y/x