In any triangle ABC, prove the following:

b sin B – c sin C = a sin (B – C)

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒c = k sin C

Similarly, b = k sin B

And a = k sin A

Here we will consider LHS, so we get

LHS = b sin B – c sin C

Substituting corresponding values in the above equation, we get

⇒ = k sin B sin B –k sin C sin C

⇒ = k (sin² B – sin² C )……….(ii)

But,

Substituting the above values in equation (ii), we get

⇒ = k(sin(B + C) sin(B - C))

But A + B + C = π⇒ B + C = π –A, so the above equation becomes,

⇒ = k(sin(π –A) sin(B - C))

But sin (π - θ) = sin θ

⇒ = k(sin(A) sin(B - C))

From sine rule, a = k sin A, so the above equation becomes,

⇒ = a sin(B - C) = RHS

Hence proved