If a, b, and c are the interior angles of triangle abc, prove that 4 sina sinb sinc = sin2a + sin2b + sin2c

NOTES:

1)⇒    A + B + C = 180°

       A + B + C = π

       A + B        = π - C

2)⇒⇒ sin (A + B) = sin (π - C)

                         = (sin π)(cos C) - (sin C)(cos π)

                          = (0)(cos C) - (sin C)(-1)

                          = 0 - (-sin C)

                          = sin C

3)⇒⇒⇒cos (A + B) = cos (π - c)

                            = (cos π)(cos C) + (sin π)(sin C)

                            = (-1)(cos C) + (0)(sin C)

                            = - cos C              

4)⇒⇒⇒⇒ sin 2A + sin 2B = 2 sin (A + B) cos (A - B)

PROOF (from left side):

sin 2A + sin 2B + sin 2C    =    4 sin A sin B sin C

2 sin (A + B) cos (A - B) + sin 2C     refer to NOTE 4

2 sin (A + B) cos (A - B) + 2 sin C cos C  double angle formula

2 sin C cos (A - B) + 2 sin C cos C   refer to NOTE 2

2 sin C [cos (A - B) + cos C]     factored out 2 sin C

2 sin C [cos (A - B) - (cos(A + B)]   refer to NOTE 3

2 sin C [2 sin A sin B]           sum/difference formula

4 sin A sin B sin C      multiplied 2 sin C by 2 sin A sin B

Proof completed: 4 sin A sin B sin C   =   4 sin A sin B sin C

  the correct choice is marked

step-by-step explanation:

the first two graphs show multiple turning points. the last graphs matches the description in every detail.