If A + B + C = π/2, prove the following (i) sin 2A + sin 2B + sin 2C

1.	If A + B + C = π/2, prove the following (i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C (ii) COS 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.
Answer» (i) LHS = (sin 2A + sin 2B) + sin 2C = 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin (90° – C) cos (A – B) + 2 sin C cos C = 2 cos C [cos (A – B) + sin C] + cos (A + B) (∴ A + B = π/2 – C) = 2 cos C [cos (A – B) + cos (A + B)] = 2 cos C [2 cos A cos B] = 4 cos A cos B cos C = RHS (ii) LHS = (cos 2A + cos 2B) + cos 2C = 2 cos (A + B) cos (A – B) + 1 – 2 sin² C = 1 + 2 sin C (cos (A – B) – 2 sin² C) {∴ cos (A + B) = cos (90° – C) = sin C} = 1 + 2 sin C [cos (A- B) – sin C] = 1 + 2 sin C [cos (A – B) – cos (A + B)] = 1 + 2 sin C [2 sin A sin B] = 1 + 4 sin A sin B sin C = RHS

If A + B + C = π/2, prove the following (i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C (ii) COS 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.

Answer»

(i) LHS = (sin 2A + sin 2B) + sin 2C

= 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin

(90° – C) cos (A – B) + 2 sin C cos C

= 2 cos C [cos (A – B) + sin C] + cos (A + B) (∴ A + B = π/2 – C)

= 2 cos C [cos (A – B) + cos (A + B)]

= 2 cos C [2 cos A cos B]

= 4 cos A cos B cos C = RHS

(ii) LHS = (cos 2A + cos 2B) + cos 2C

= 2 cos (A + B) cos (A – B) + 1 – 2 sin² C

= 1 + 2 sin C (cos (A – B) – 2 sin² C)

{∴ cos (A + B) = cos (90° – C) = sin C}

= 1 + 2 sin C [cos (A- B) – sin C]

= 1 + 2 sin C [cos (A – B) – cos (A + B)]

= 1 + 2 sin C [2 sin A sin B]

= 1 + 4 sin A sin B sin C = RHS