Need to prove: a sin A – b sin B = c sin (A – B)

Left hand side,

a sin A – b sin B

(b cosC + c cosB) sinA – (c cosA + a cosC) sinB

b cosC sinA + c cosB sinA – c cosA sinB – a cosC sinB

c(sinA cosB – cosA sinB) + cosC(b sinA – a sinB)

We know that, where R is the circumradius.

Therefore,

c(sinA cosB – cosA sinB) + cosC(2R sinB sinA – 2R sinA sinB)

c(si nA cosB – cosA sinB)

c sin (A – B)

Right hand side. [Proved]