[กิตติ]

จาก$sin3A=3sinA-4sin^3A \rightarrow sin^3A= \dfrac{3sinA-sin3A}{4} $
$sinA+sinB = 2sin(\frac{A+B}{2} )cos(\frac{A-B}{2})$
และ$cosA+cosB=2cos(\frac{A+B}{2} )cos(\frac{A-B}{2})$
และ$A+B+C=\pi \rightarrow \frac{A}{2}+\frac{B}{2}+\frac{C}{2} =\frac{\pi }{2}$
$\frac{C}{2} =\frac{\pi }{2}-(\frac{A}{2}+\frac{B}{2})$
$\frac{3C}{2} =\frac{3\pi }{2}-(\frac{3A}{2}+\frac{3B}{2})$

$sin^3A+sin^3B+sin^3C = \dfrac{1}{4}\left[\,\right.3(sinA+sinB+sinC)-(sin3A+sin3B+sin3C)\left.\,\right] $

$sinA+sinB+sinC = (2sin(\frac{A+B}{2} )cos(\frac{A-B}{2}))+2sin(\frac{C}{2} )cos(\frac{C}{2})$
$=2cos(\frac{C}{2})cos(\frac{A-B}{2})+2sin(\frac{C}{2} )cos(\frac{C}{2})$
$=2cos\frac{C}{2}(cos(\frac{A-B}{2})+sin(\frac{C}{2} ))$
$=2cos\frac{C}{2}(cos(\frac{A-B}{2})+cos(\frac{A+B}{2} ))$
$=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$

$sin3A+sin3B+sin3C =\left[\,\right. 2sin(\frac{3A+3B}{2} )cos(\frac{3A-3B}{2} )\left.\,\right] +(2sin\frac{3C}{2}cos\frac{3C}{2})$
$sin(\frac{3A+3B}{2} )= -cos\frac{3C}{2} ,sin\frac{3C}{2}= -cos(\frac{3A+3B}{2} )$
$sin3A+sin3B+sin3C = 2cos\frac{3C}{2}\left[\,\right. sin\frac{3C}{2}-cos(\frac{3A-3B}{2})\left.\,\right] $
$=2cos\frac{3C}{2}\left[\,\right. -cos(\frac{3A+3B}{2} )-cos(\frac{3A-3B}{2})\left.\,\right] $
$= -2cos\frac{3C}{2}\left[\,\right. cos(\frac{3A+3B}{2} )+cos(\frac{3A-3B}{2})\left.\,\right]$
$ = -4cos\frac{3A}{2}cos\frac{3B}{2}cos\frac{3C}{2}$

$sin^3A+sin^3B+sin^3C = \dfrac{1}{4}\left[\,\right.3(sinA+sinB+sinC)-(sin3A+sin3B+sin3C)\left.\,\right] $
$=\dfrac{1}{4}\left[\,\right.3(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2})-(-4cos\frac{3A}{2}cos\frac{3B}{2}cos\frac{3C}{2})\left.\,\right] $
$=3cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{3A}{2}cos\frac{3B}{2}cos\frac{3C}{2}$

$sin^3A+sin^3B+sin^3C = 3cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{3A}{2}cos\frac{3B}{2}cos\frac{3C}{2}$
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