Question

prove that

a Sin (B - C) + b Sin (C - A) + c Sin (A - B) = 0

Answer 1

$\bold{\huge{Hay!!}}$


$\bold{Dear\:user!!}$



$\bold{\underline{Question-}}$

prove that

a Sin (B - C) + b Sin (C - A) + c Sin (A - B) = 0


$\bold{\underline{Answer-}}$

$let \: \: \: \frac{a}{sin \: A} = \frac{b}{sin \: B} = \frac{c}{sin \: C} = k$


Now,

$a \: = k \: sin \: A \\ b = k \: sin \: B \\ c = k \: sin \: C$


LHS, =


$= k[sin \: A \: sin \: (B - C) + sin \:B \: sin \: (C - A ) + sin \: C \: sin \: (A - B)] \\ \\ \\ = k[sin \: (B + C)sin \: (B - C) \: + \: sin \: (C + A) \: sin \:( C - A) + sin \: (A + B) \: sin \: (A - B)] \\ \\ \\ \\ = k[( {sin}^{2} B - {sin}^{2} C) + ( {sin}^{2} C - {sin}^{2} A) + ( {sin}^{2} A - {sin}^{2} B)] \\ \\ \\ \\ = k \times 0 = 0$



LHS = RHS



Hence, proved

Answer 2

Answer:

Step-by-step explanation:

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