Cho tam giác ABC, chứng minh:

$sin\widehat A + \sin \widehat B + \sin \widehat C = 4.\cos \frac{{\widehat A}}{2}.\cos \frac{{\widehat B}}{2}.\cos \frac{{\widehat C}}{2}$.

Ta có: $\widehat A + \widehat B + \widehat C = 180^\circ$

$\Rightarrow \sin \frac{{(\widehat A + \widehat B)}}{2} = \sin \left( {\frac{{180^\circ }}{2} - \frac{{\widehat C}}{2}} \right) = \cos \frac{{\widehat C}}{2}$

Tương tự ta có: $\sin \frac{{\widehat C}}{2} = \cos \frac{{\widehat A + \widehat B}}{2}$

$\Rightarrow \sin \widehat A + \sin \widehat B + \sin \widehat C = 2\cos \frac{{\widehat C}}{2}.\cos \frac{{\widehat A - \widehat B}}{2} + 2\cos \frac{{\widehat A + \widehat B}}{2}.\cos \frac{{\widehat C}}{2}$

$= 2\cos \frac{{\widehat C}}{2}\left( {\cos \frac{{\widehat A - \widehat B}}{2} + \cos \frac{{\widehat A + \widehat B}}{2}} \right)$

$= 4.\cos \frac{{\widehat A}}{2}.\cos \frac{{\widehat B}}{2}.\cos \frac{{\widehat C}}{2}$ (đpcm).