Question

if A + B + C = 180 then prove that sin3a + sin3b + sin3c = - 4cos3a/2*cos3b/2*cos3c/2

Answer 1

given, A + B + C = 180°
so, 3A + 3B + 3C = 540°

LHS = sin3A + sin3B + sin3C

= 2sin(3A + 3B)/2. cos(3A - 3B)/2 + sin3C

= 2sin(540° - 3C)/2 . cos(3A - 3B)/2 + sin3C

= 2sin(270° - 3C/2) . cos(3A - 3B)/2 + 2sin3C/2. cos3C/2

= -2cos3C/2 . cos(3A - 3B)/2 + 2cos3C/2 . sin3C/2

= -2cos3C/2 [ cos(3A - 3B)/2 - sin3C/2 ]

= -2cos3C/2 [ cos(3A - 3B)/2 + cos(270° - 3C/2)]

= -2cos3C/2 [ 2cos(3A/2- 3B/2 + 270° - 3C/2)/2. cos(3A/2 - 3B/2 - 270° + 3C/2)/2]

= -4cos3C/2 [ cos{3A/2 + 270° - (3B + 3C)/2}/2.cos{(3A + 3C)/2 - 3B/2 - 270°}/2 ]

= -4cos3C/2 [ cos(3A/2 + 3A/2)/2 . cos(- 3B/2 - 3B/2)/2 ]

= -4cos3A/2 . cos3B/2 cos3C/2 = RHS

Answer 2

HELLO DEAR,




given, A + B + C = 180°-------( 1 )

now,
sin3A + sin3B + sin3C

=> 2sin(3A + 3B)/2. cos(3A - 3B)/2 + sin3C
[as, 3A + 3B + 3C = 540° ----from( 1 )]

=> 2sin(540° - 3C)/2 . cos(3A - 3B)/2 + sin3C

=> 2sin(270° - 3C/2) . cos(3A - 3B)/2 + 2sin3C/2. cos3C/2

=> -2cos3C/2 . cos(3A - 3B)/2 + 2cos3C/2 . sin3C/2

=> -2cos3C/2 [ cos(3A - 3B)/2 - sin3C/2 ]

=> -2cos3C/2 [ cos(3A - 3B)/2 + cos(270° - 3C/2)]

=> -2cos3C/2 [ 2cos(3A/2- 3B/2 + 270° - 3C/2)/2. cos(3A/2 - 3B/2 - 270° + 3C/2)/2]

=> -4cos3C/2 [ cos{3A/2 + 270° - (3B + 3C)/2}/2.cos{(3A + 3C)/2 - 3B/2 - 270°}/2 ]

=> -4cos3C/2 [ cos(3A/2 + 3A/2)/2 . cos(- 3B/2 - 3B/2)/2 ]

=> -4cos3A/2 . cos3B/2 . cos3C/2



I HOPE IT'S HELP YOU DEAR,
THANKS