Question

A, B, C are the angles of a triangle. Prove sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
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Answer 1

Answer:

Step-by-step explanation:

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

=2 sin C (cos( A−B ) + cos C ), since sin (A+B) = sin { 180 - (A + B)} = sin C

=4 sin C cos 1/2 (A+C−B )cos 1/2  (A−B−C)

=4 sin C  cos 1/2  ([180−B]−B)  cos 1/ 2  (A−[180−A] )

=4 sin  C cos (90−B )cos  (A−90)

=4 sin C cos (90−B )cos  (90−A)

=4 sin A sin Bsin C

Proved