In any ΔABC, prove that 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

Need to prove: a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0

From left hand side,

a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B)

a²((1 - sin²B) – (1 - sin²C)) + b²((1 - sin²C) – (1 - sin²A)) + c²((1 - sin²A) – (1 - sin²B))

a²( - sin²B + sin²C) + b²( - sin²C + sin²A) + c²( - sin²A + sin²B)

We know that, where R is the circumradius.

Therefore,

a = 2R sinA ---- (a)

Similarly, b = 2R sinB and c = 2R sinC

So,

4R²[ sin²A( - sin²B + sin²C) + sin²B( - sin²C + sin²A) + sin²C( - sin²A + sin²B)

4R²[ - sin²Asin²B + sin²Asin²C – sin²Bsin²C + sin²Asin²B – sin²Asin²C + sin²Bsin²C ]

4R² [0]

0 [Proved]