Question

In a triangle ABC, prove that:
(i) a sin (B-C) + b sin (C - A) + C sin (A-B) = 0​

Answer 1

PROOF:

sin A/a = sin B/b = sin C/c = a constant k (Law of sines)

∴ sin A/a = k

   sin A = ak

Similarly, sin B/b = k

  sin B = bk

Similarly, sin C/c = k

  sin C = ck

a sin (B - C) = a (sin B cos C - cos B sin C)

                 =  a (bk cos C - cos B ck)  [Substituting sin B = bk, sin C = ck]

                 =  ak (b cos C - c cos B)

b sin (C - A) = b (sin C cos A - cos C sin A)

                  = b (ck cos A - cos C ak)  [Substituting sin C = ck, sin A = ak]

                  =bk (c cos A - a cos C)

c sin (A - B) = c (sin A cos B - cos A sin B)

                 = c (ak cos B - cos A bk)  [Substituting sin A = ak, sin B = bk]

                 = ck (a cos B - b cos A)

∴ L.H.S. = a sin (B - C) + b sin (C - A) + c sin (A - B)

        = ak (b cos C - c cos B) + bk (c cos A - a cos C) + ck (a cos B - b cos A)

        = abk cos C - ack cos B + bck cos A - abk cos C + ack cos B - bck cos A

        = abk cos C - abk cos C + ack cos B - ack cos B + bck cos A - bck cos A

        = 0

        = R.H.S.

                         

                                              HENCE PROVED