a) Áp dụng tổng các góc trong ∆ABC có:

$\widehat A + \widehat B + \widehat C = {180^o}$

$\Rightarrow \widehat A + \widehat B = {180^o} - \widehat C$

$\Leftrightarrow \frac{{\widehat A + \widehat B}}{2} = \frac{{{{180}^o} - \widehat C}}{2} = {90^o} - \frac{{\widehat C}}{2}$

$\Rightarrow \sin \left( {\frac{{\widehat A + \widehat B}}{2}} \right) = \sin \left( {{{90}^o} - \frac{{\widehat C}}{2}} \right) = \cos \frac{{\widehat C}}{2}$;

b) Ta có: $\widehat A + \widehat B + \widehat C = {180^o}$

$\tan \left( {2\widehat A + \widehat B + \widehat C} \right) = \tan \left( {\widehat A + \widehat A + \widehat B + \widehat C} \right)$

$= \tan \left( {\widehat A + {{180}^o}} \right) = \tan \widehat A$

$\Rightarrow \tan \left( {2\widehat A + \widehat B + \widehat C} \right) = \tan \widehat A$

c) Ta có: $\widehat A + \widehat B + \widehat C = {180^o}$

$\Leftrightarrow \frac{{\widehat A + \widehat B + 3\widehat C}}{2} = \frac{{{{180}^o} + 2\widehat C}}{2} = {90^o} - \widehat C$

$\Rightarrow \sin \left( {\frac{{\widehat A + \widehat B + 3\widehat C}}{2}} \right) = \sin \left( {{{90}^o} - \widehat C} \right) = \cos \widehat C$