Question

Prove A +B+C = π, prove that

Answer 1

Step-by-step explanation:

Solution :

LHS = cos 2A + cos 2B + cos 2C

= 2cos (A+B) cos (A–B) + 2cos²C – 1

= 2cos (π–C) cos (A–B) + 2cos²C – 1

= –2cos C cos (A–B) + 2cos²C – 1

= –2cos C [cos (A–B) – cos C] –1

= –1–2cos C [cos(A–B)–cos{π–(A+B)}]

= –1–2cos C [cos (A–B) + cos (A+B)]

= –1–2cos C [2cos A cos B]

= –1–4cos A cos B cos C = RHS.

Therefore :

cos 2A + cos 2B + cos 2C = –1–4cos A cos B cos C.