If A + B + C = 180°, then prove that sin 2A + sin 2B + sin 2C = 4 sin

1.	If A + B + C = 180°, then prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
Answer» sin 2A + sin 2B + sin 2C = 2 sin C. cos (A – B) + 2 sin C. cos C = 2 sin C (cos(A – B) + cos C] = 2 sin C[cos (A – B) + cos {180 – (A + B)}] = 2 sin C[cos (A – B) – cos (A + B)] = 2 sin C[-2 sin A . sin(- B)] = 4 sin A. sin B.sin C

If A + B + C = 180°, then prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Answer»

sin 2A + sin 2B + sin 2C

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

= 2 sin C (cos(A – B) + cos C]

= 2 sin C[cos (A – B) + cos {180 – (A + B)}]

= 2 sin C[cos (A – B) – cos (A + B)]

= 2 sin C[-2 sin A . sin(- B)]

= 4 sin A. sin B.sin C

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If A + B + C = 180°, then prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

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