In any ΔABC, prove that a sin A – b sin B = c sin (A – B)

1.	In any ΔABC, prove that a sin A – b sin B = c sin (A – B)
Answer» 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 a/sinA = b/sinB = c/sinC = 2R 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]

Answer»

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 a/sinA = b/sinB = c/sinC = 2R 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]

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