Question

Prove the following:

In Δ ABC, A + B + C = π, show that cos2A + cos2B + cos2C = – 1 – 4 cosA cosB cosC

Answer 1

L.H.S. = cos 2A + cos 2B + cos 2C

= 2 . cos(\(\frac{2\,A+2\,B}{2}\)).cos(\(\frac{2\,A-2\,B}{2}\)) + cos 2C

= 2.cos(A + B).cos (A – B) + 2cos² C – 1 

In ΔABC, A + B + C = π 

∴ A + B = π – C 

∴ cos(A + B) = cos(π – C) 

∴ cos(A + B) = – cos C ………….(i)

∴ L.H.S. = – 2.cos C. cos (A – B) + 2.cos² C – 1 … [From(i)] 

= – 1 – 2.cosC.[cos(A – B) – cos C] 

= – 1 – 2.cos C.[cos(A – B) + cos(A + B)]… [From (i)] 

= – 1 – 2.cos C.(2.cos A. cos B) 

= – 1 – 4.cos A. cos B. cos C = R.H.S.