In any triangle ABC, prove the following: a 2 (cos 2 B – cos 2 C) + b 2 (cos 2 C – cos 2 A) + c 2 (cos 2 A – cos 2 B) = 0

Question

Philoid · Accepted Answer

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

⇒ a = k sin A, b = k sin B, c = k sin C

Here we will consider LHS, so we get

LHS = a² (cos² B – cos² C) + b² (cos² C – cos² A) + c² (cos² A – cos² B)

Substituting corresponding values from sine rule in above equation, we get

= (k sin A)² (cos² B – cos² C) + (k sin B)² (cos² C – cos² A) + (k sin C)² (cos² A – cos² B)

= k²(sin²A cos² B – sin²A cos² C + sin²B cos² C – sin²B cos² A + sin²A cos² A – sin²A cos² B)

Cancelling the like terms, we get

LHS = 0 = RHS

Hence proved