In a ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.

1.	In a ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.
Answer» LHS = a cos A+ 6 cos B + c cos C Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k = (k/2) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C] = (k/2) [sin 2A + sin 2B + sin 2C] = (k/2) [2 sin (A + B) . cos (A – B) + 2 sin C . cos C] = (k/2) [2 sin (A – B) . cos (A – B) + 2 sin C . cos C] = (k/2) [2 sin C . cos (A – B) + 2 sin C . cos C] = k sin C [cos(A – B) + cos C] = k sin C[cos(A - B) + cos(π - bar(A + B)) = k sin C [cos (A – B) – cos (A + B)] = k sin C . 2 sin A sin B = 2k sin A . sin B sin C = 2a sin B sin C = RHS

Answer»

LHS = a cos A+ 6 cos B + c cos C

Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k

= (k/2) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]

= (k/2) [sin 2A + sin 2B + sin 2C]

= (k/2) [2 sin (A + B) . cos (A – B) + 2 sin C . cos C]

= (k/2) [2 sin (A – B) . cos (A – B) + 2 sin C . cos C]

= (k/2) [2 sin C . cos (A – B) + 2 sin C . cos C]

= k sin C [cos(A – B) + cos C]

= k sin C[cos(A - B) + cos(π - bar(A + B))

= k sin C [cos (A – B) – cos (A + B)]

= k sin C . 2 sin A sin B

= 2k sin A . sin B sin C

= 2a sin B sin C = RHS

Discussion