Cho tam giác ABC, chứng minh: sin góc A + sin góc B + sin góc C = 4cos góc A/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).