Question

in triangle abc prove that asin(B-C)+bsin(C-A)+csin(A-B)=0

Answer 1

Use sine formula ,
$\bold{\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k}$
∴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

Answer 2

Solution :-

sinA = ak

sinB = bk

sinC = ck

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

= akcosB - cosA.bk

= K(acosB - bcosA}

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