Question

In any A ABC prove the following:
(i) asin A - bsin B=csin(A-B)​

Answer 1

Answer:

Step-by-step explanation:

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