Question

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.
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Answer 1

Answer:

Step-by-step explanation:

Also sin(A - B) = sinA.cosB - cosA.sinB

= akcosB - cosA.bk

= K(acosB - bcosA}

Similarly, sin(B - C) = k(bcosC - ccosB)

sin(C - A) = k(ccosA - acosC)

LHS = asin(B- C) + bsin(C - A) + csin(A - B)

= ak(bcosC - ccosB) + bk(acosC - ccosA) + ck(acosB - bcosA)

= k(bccosA - bccosA) + k(accosB - accosB) + k(abcosC - abcosC)

= 0 + 0 + 0 = 0 = RHS

Answer 2

Answer:

Use sine formula ,

∴sinA = ak

sinB = bk

sinC = ck

Also sin(A - B) = sinA.cosB - cosA.sinB

= akcosB - cosA.bk

= K(acosB - bcosA}

Similarly, sin(B - C) = k(bcosC - ccosB)

sin(C - A) = k(ccosA - acosC)

LHS = asin(B- C) + bsin(C - A) + csin(A - B)

= ak(bcosC - ccosB) + bk(acosC - ccosA) + ck(acosB - bcosA)

= k(bccosA - bccosA) + k(accosB - accosB) + k(abcosC - abcosC)

= 0 + 0 + 0 = 0 = RHS

Step-by-step explanation:

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