Buktikan bahwa : sin2A+sin2B+sin2C = 4sinAsinBsinC plis tolong banget
whongaliemA + B + C = 180° C = 180° - (A + B) 2C = 360° - 2 (A + B) sin 2C = sin [360 - 2 (A + B)} = - sin 2(A + B) = - 2 sin (A + B) .cos (A + B)
sin 2A + sin 2B + sin 2C = 4.sin A .sin B .sin 2C
(sin 2A + sin 2B)+ sin 2C = 2{.sin 1/2 (2A + 2B).cos 1/2 (2A -2 B)}+sin 2C = 2.{sin (A + B).cos (A - B } - 2sin (A + B).cos (A + B) = 2.sin (A + B) { cos (A - B) - (cos A + B) } = 2.sin (A + B){ - 2 sin 1/2 (A - B + A + B).sin 1/2(A - B - A - B) = 2 sin (180 - C) { - 2sin 1/2 (2A) .sin 1/2(-2B)} = 2 sin C ( - 2 sin A . - sin B) = 4.sin A .sin B . sin C .... (terbukti)
C = 180° - (A + B)
2C = 360° - 2 (A + B)
sin 2C = sin [360 - 2 (A + B)}
= - sin 2(A + B)
= - 2 sin (A + B) .cos (A + B)
sin 2A + sin 2B + sin 2C = 4.sin A .sin B .sin 2C
(sin 2A + sin 2B)+ sin 2C = 2{.sin 1/2 (2A + 2B).cos 1/2 (2A -2 B)}+sin 2C
= 2.{sin (A + B).cos (A - B } - 2sin (A + B).cos (A + B)
= 2.sin (A + B) { cos (A - B) - (cos A + B) }
= 2.sin (A + B){ - 2 sin 1/2 (A - B + A + B).sin 1/2(A - B - A - B)
= 2 sin (180 - C) { - 2sin 1/2 (2A) .sin 1/2(-2B)}
= 2 sin C ( - 2 sin A . - sin B)
= 4.sin A .sin B . sin C .... (terbukti)